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OSMANIA UNIVERSITY LIBRARY, 

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INTERNATIONAL CHEMICAL SERIES 
Louis P. HAMMETT, PH.D., Consulting Editor 

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CALCULATIONS OF 
ANALYTICAL CHEMISTRY 



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CALCULATIONS OF 
ANALYTICAL CHEMISTRY 

FORMERLY PUBLISHED UNDER THE TITLE 

Calculations of Quantitative Chemical Analysis 



by LEICESTER F. HAMILTON, S.B. 

Professor of Analytical Chemistry 
Massachusetts Institute of Technology 

and STEPHEN G. SIMPSON, PH.D. 

Associate Professor of Analytical Chemistry 
Massachusetts Institute of Technology 



Fourth Edition 
Third Impression 



McGRAW-HILL BOOK COMPANY, ING. 

NEW YORK AND LONDON 
1947 



CALCULATIONS OF ANALYTICAL CHEMISTRY 

Formerly published under the title 
Calculations of Quantitative Chemical Analysis 

COPYRIGHT, 1922, 1927, 1939, 1947, BY THE 
McGRAw-HiLL BOOK COMPANY, INC. 

PRINTED IN THE UNITED STATES OF AMERICA 

All rights reserved. This book, or 

parts thereof, may not be reproduced 

w, any form without permission of 

the publishers. 



PREFACE 

The title of this book has been clfanged from Calculations of 
Quantitative Chemical Analysis to Calculations of Analytical Chem- 
istry because the subject matter has been expanded to cover the 
stoichiometry of both qualitative and quantitative analysis. 

In order to include calculations usually covered in courses in 
qualitative analysis, some rearrangements of material have been 
made, new sections have been added, and chapters dealing with 
equilibrium constants and with the more elementary aspects of 
analytical .calculations have been considerably expanded. Al- 
together, the number of sections has been increased from 78 to 
114 and the number of problems from 766 to 1,032. 

The greater part of the book is still devoted to the calculations 
of quantitative analysis. Short chapters on conductometric and 
amperometric titrations and a section on calibration of weights 
have been added, and many other changes and additions have 
been made at various points in the text. A section reviewing the 
use of logarithms has been inserted, and a table of molecular 
weights covering most of the problems in the book is included in 
the Appendix. 

It is felt that every phase of general analytical chemistry is 
adequately covered by problems, both with and without answers, 
and that most of the problems require reasoning on the part of the 
student and are not solved by simple substitution in a formula. 

LEICESTER F. HAMILTON 
STEPHEN G. SIMPSON 
CAMBRIDGE, MASS., 
February, 1947. 



CONTENTS 



PREFACE v 

PART I. GENERAL ANALYSIS 

CHAPTER I. MATHEMATICAL, OPERATIONS 

1. Factors Influencing the Reliability of Analytical Results 1 

2. Deviation Measures as a Means of Expressing Reliability ... . 2 

3. Significant Figures as a Means of Expressing Reliability 3 

4. Rules Governing the Use of Significant Figures in Chemical Com- 
putations 3 

5. Conventions Regarding the Solution of Numerical Problems .... 6 
Problems 1-18 7 

6. Rules Governing the Use of Logarithms .... 9 

7. Method of Using Logarithm Tables . . 13 

8. Use of the Slide Rule 14 

Problems 19-24 15 

CHAPTER II. CHEMICAL, EQUATIONS 

9. Purpose of Chemical Equations 16 

10. Types of Chemical Equations 16 

11. lonization of Acids, Bases, and Salts 17 

12. Ionic Equations Not Involving Oxidation 18 

13. Oxidation Number 20 

14. Ionic Oxidation and Reduction Equations 21 

Problems 25-43 24 

CHAPTER III. CALCULATIONS BASED ON FORMULAS AND EQUATIONS 

15. Mathematical Significance of a Chemical P^ormula . 28 

16. Formula Weights 28 

17. Mathematical Significance of a Chemical Equation 29 

Problems 44-70 32 

CHAPTER IV. CONCENTRATION OF ^ SOLUTIONS 

18. Methods of Expressing Concentration 36 

19. Grains per Unit Volume 3f> 

vii 



CONTENTS 

20. Percentage Composition. . . . . 36 

21. Specific Gravity 36 

22. Volume Ratios 37 

23. Molar and Formal Solutions 37 

24. Equivalent Weight and Normal Solution 38 

25. Simple Calculations Involving Equivalents, Milliequivalents, and 

Normality 39 

Problems 71-86 43 

CHAPTER V. P]quiLiBRiUM CONSTANTS 

26. Law of Mass Action 46 

27. Ion Product Constant of Water 47 

28. pll Value 48 

Problems 87-94 49 

29. lonization Constant 50 

30. Common Ion Effect. Buffered Solution 52 

31. lonization of Polybasic Acids 53 

32. Activity and Activity Coefficients 55 

33. Dissociation Constants of Complex Ions 56 

Problems 95-128 57 

34. Solubility Product 60' 

35. Fractional Precipitation 62 

Problems 129-159 64 

36. Application of Buffered Solutions in Analytical Chemistry 67 

37. Control of Acidity in Hydrogen Sulficie Precipitations 68 

38. Separations by Means of Complex-ion Formation 69 

Problems 160-178 71 

39. Distribution Ratio 73 

Problems 179-185 74 

CHAPTER VI. OXIDATION POTENTIALS 

40. Relation of the Electric Current to Oxidation-reduction ("Redox") 
Reactions 76 

41. Specific Electrode Potentials 76 

42. Rules for Writing Equations for Half-cell Reactions 79 

43. Oxidation-reduction Equations in Terms of Half-cell Reactions. . . 80 

44. Relation between Electrode Potential and Concentration 83 

45. Calculation of the Extent to Which an Oxidation-reduction Reaction 
Takes Place 86 



CONTENTS ix 

46. Calculation of Equilibrium Constant from Electrode Potentials . . 87 
Problems 186-218 88 

PART II. GRAVIMETRIC ANALYSIS 
CHAPTER VII. THE CHEMICAL, BALANCE 

47. Sensitiveness of the Chemical Balance 93 

48. Method of Swings 94 

49. Conversion of Weight in Air to Weight in Vacuo 95 

50. Calibration of Weights 97 

Problems 219-240 99 

CHAPTER VIII. CALCULATIONS OF GRAVIMETRIC ANALYSIS 

51 . Law of Definite Proportions Applied to Calculations of Gravimetric 
Analysis 102 

52. Chemical Factors 104 

53. Calculation of Percentages 105 

Problems 241-266 106 

54. Calculation of Atomic Weights 109 

Problems 267-275 109 

55. Calculations Involving a Factor Weight Sample 110 

Problems 276-284 112 

56. Calculation of the Volume of a Reagent Required for a Given Reaction 1 13 
Problems 285-304 116 

57. Indirect Analyses 118 

Problems 305-331 121 

CHAPTER IX. ELECTROLYTIC METHODS 

58. Decomposition Potential 125 

59. Analysis by Electrolysis 126 

Problems 332-359 130 

CHAPTER X. CALCULATIONS FROM REPORTED PERCENTAGES 

60. Calculations Involving the Elimination or Introduction of a Con- 
stituent 135 

61. Cases Where Simultaneous Volatilization and Oxidation or Reduction 

Occur 136 

Problems 360-380 138 

62. Calculation of Molecular Formulas from Chemical Analyses .... 141 

63. Calculation of Empirical Formula of a Mineral 144 



x CONTENTS 

64. Calculation of Formulas of Minerals Exhibiting Isomorphic Re- 
placement 145 

Problems 381-424 147 

PART III. VOLUMETRIC ANALYSIS 
CHAPTER XI. CALIBRATION OF MEASURING INSTRUMENTS 

65. Measuring Instruments in Volumetric Analysis 153 

66. Calculation of True Volume 153 

Problems 425-435 155 

CHAPTER XII. NEUTRALIZATION METHODS (AOIDIMETRY 
AND ALKALIMETRY) 

67. Divisions of Volumetric Analysis 158 

68. Equivalent Weights Applied to Neutralization Methods 158 

Problems 436-456 160 

69. Normality of a Solution Made by Mixing Similar Components . . 161 
Problems 457-466 162 

70. Volume-normality-milliequivalent Relationship 163 

71. Adjusting Solution to a Desired Normality 163 

Problems 467-480 164 

72. Volume and Normality Relationships between Reacting Solutions . 166 
Problems 481-492 166 

73. Determination of the Normality of a Solution 168 

74. Conversion of Data to Milliequivalents 169 

Problems 493-514 170 

75. Calculation of Percentage Purity from Titratioii Values 173 

76. Volumetric Indirect Methods 176 

Problems 515-532 177 

77. Problems in Which the Volume of Titrating Solution Bears a Given 

Relation to the Percentage 180 

Problems 533-545 181 

78. Determination of the Proportion in Which Components Are Present 

in a Pure Mixture 183 

79. Analysis of Fuming Sulfuric Acid 184 

Problems 546-564 186 

80. Indicators 188 

81. Equivalence Point 189 

82. Determination of pH Value at the Equivalence Point 191 

83. Calculation of the Degree of Hydrolysis of a Salt 196 



CONTENTS xi 

Problems 565-591 196 

84. Titration of Sodium Carbonate 199 

85. Analyses Involving the Use of Two Indicators 199 

86. Relation of Titration Volumes to Composition of Sample 204 

Problems 592-606 205 

87. Analysis of Phosphate Mixtures 207 

Problems 607-612 209 

CHAPTER XIII. OXIDATION AND REDUCTION ("REDOX") METHODS 

(OXIDIMETRY AND REDUCTIMETRY) 

88. Fundamental Principles 211 

89. Equivalent Weights of Oxidizing and Reducing Agents 211 

90. Calculations of Oxidation and Reduction Processes 216 

Problems 613-630 217 

91. Permanganate Process 219 

92. Bichromate Process 225 

93. Ceric Sulfate or Cerate Process 226 

Problems 631-682 227 

94. lodimetric Process 234 

Problems 683-712 238 

CHAPTER XIV. PRECIPITATION METHODS (PRECIPITIMETRY) 

95. Equivalent Weights in Precipitation Methods 243 

Problems 713-734 245 

CHAPTER XV. COMPLEX-ION FORMATION METHODS (COMPLEXIMETRY) 

96. Equivalent Weights in Complex-ion Methods 249 

Problems 735-751 252 

PART IV. ELECTROMETRIC METHODS 

CHAPTER XVI. POTENTIOMETRIC TITRATIONS 

97. Potentiometric Acidimetric Titrations 255 

98. Simple Potentiometric Titration Apparatus 256 

99. Quinhy drone Electrode 257 

100. Glass Electrode 258 

101. Potentiometric "Redox" Titrations 259 

102. Potentiometric Precipitation Titrations 261 

Problems 752-775 261 



xii CONTENTS 

CHAPTER XVII. CONDUCTOMETRIC TITRATIONS 

103. Conductance 266 

104. Mobility of Ions 266 

105. Conductometric Acidimetric Titrations 267 

106. Conductometric Precipitation Titrations 270 

107. Conductometric Titration Apparatus 271 

Problems 776-789 272 

CHAPTER XVIII. AMPEROMETRIC TITRATIONS 

108. Principle of an Amperometric Titration 275 

Problem 790 278 

PART V. GAS ANALYSIS 
CHAPTER XIX. CALCULATIONS OF GAS ANALYSES 

109. Fundamental Laws 279 

110. Gas-volumetric Methods 281 

111. Correction for Water Vapor 282 

112. Calculations of Gas- volumetric Analyses 283 

Problems 791-813 284 

113. Absorption Methods 286 

114. Combustion Methods 287 

Problems 814-836 292 

PART VI. COMMON ANALYTICAL DETERMINATIONS 

PART vii. "PROBLEMS ON SPECIFIC GROUPS AND 

DETERMINATIONS 

A. Qualitative Analysis 

Silver Group 309 

Hydrogen Sulfide Group 310 

Ammonium Sulfide Group 311 

Alkaline Earth and Alkali Groups 312 

Anion Groups 313 

B. Quantitative Analysis 

Water 313 

Sodium, Potassium 313 

Ammonium, Ammonia, Nitrogen 314 

Silver, Mercury, Gold, Platinum 315 

Halogens, Cyanide, Thiocyanate, Halogen acids 315 



CONTENTS xiii 

Barium, Strontium, Calciuirif Magnesium 163 

Limestone, Lime, Cement 317 

Iron, Aluminum, Titanium 317 

Cerium, Thorium, Zirconium, Uranium, Beryllium, Bismuth, Boron . . 320 

Copper, Lead, Zinc, Cadmium, Brass 321 

Tin, Antimony, Arsenic, Bronze 323 

Carbon, Carbon Dioxide, Silicon, Tungsten, Molybdenum 324 

Chromium, Vanadium 325 

Manganese 327 

Cobalt, Nickel 329 

Phosphorus 329 

Sulfur, Selenium 331 

General and Miscellaneous Analyses 335 

APPENDIX 365 

INDEX 379 



PART I 
GENERAL ANALYSIS 

CHAPTER I 
MATHEMATICAL OPERATIONS 

1. Factors Influencing the Reliability of Analytical Results. 

Analytical chemistry is ordinarily divided into qualitative analysis 
and quantitative analysis. A compound or mixture is analyzed by 
qualitative analysis to determine what constituents or components 
are present; a compound or mixture is analyzed by quantitative 
analysis to determine the proportions in which the constituents 
or components are present. 

Calculations in qualitative analysis are limited mostly to those 
pertaining to equilibrium constants and simple weight and volume 
relationships. Calculations in quantitative analysis are more ex- 
tensive and are based upon numerical data obtained by careful 
measurement of masses and volumes of chemical substances. From 
the numerical data obtained from these measurements the desired 
proportions can be calculated. It is found; however, that duplicate 
analyses of the same substance, even when made by experienced 
analysts following identical methods, rarely give numerical values 
which are exactly the same. Furthermore, the discrepancy be- 
tween results is found to depend upon the method used, and an 
analytical result obtained by one procedure may differ appre- 
ciably from a similar result obtained by an entirely different pro- 
cedure. The most important factors which thus influence the 
precision of analytical results are the following: (1) the manipu- 
lative skill of the analyst; (2) the experimental errors of the pro- 
cedure itself, such as the slight solubility of substances assumed 
to be insoluble or the contamination of precipitates assumed to 
be pure; (3) the accuracy of the measuring instruments used; and 
(4) fluctuations of temperature and barometric pressure. In order, 
therefore, that a numerical result obtained from chemical meas- 
urements may be of scientific or technical value, the observer 
should have at least a general idea of its reliability. 

l 



2 CALCULATIONS OF ANALYTICAL CHEMISTRY 

In this connection, there should be kept in mind a distinction 
between accuracy and reliability. The accuracy of a numerical 
result is the degree of agreement between it and the true value; 
the reliability or precision of a numerical result is the degree of 
agreement between it and other values obtained under substan- 
tially the same conditions. Thus, suppose duplicate determina- 
tions of the percentage of copper in an ore gave 52.30 per cent 
and 52.16 per cent, and suppose the actual percentage was 52.32. 
It can be assumed that the analyst would report the mean or 
average of the two values obtained, namely 52.23 per cent. This 
differs from the true value by 0.09 per cent, which represents the 
absolute error of the analysis. Expressed in parts per thousand, 
the error would be 0.09/52.32 X 1,000 =1.7 parts per thousand. 
This is known as the relative error of the analysis. 

Since in most chemical analyses the true value is not known, 
it follows that the accuracy of a given determination is seldom 
known. We can speak only of the precision or reliability of the 
numerical results obtained. 

2. Deviation Measures as a Means of Expressing Reliability. 
The numerical measure of the reliability of a result is known as 
its precision measure. A type of precision measure which is of 
particular importance in careful physical and chemical work is 
the deviation measure. Suppose, for example, repeated independent 
readings of a buret gave the following values: 

(a) 43.74 (/) 43.75 

(6) 43.76 (p) 43.75 

(c) 43.76 (h) 43.76 

(d) 43.75 (i) 43.73 

(e) 43.77 

The most probable value tor this reading is obviously the mean, 
43.753, which is obtained by dividing the sum of the readings by 
the number of readings taken. The deviation of each measurement 
from this mean, regardless of sign, is shown in the following: 

(a) 0.013 (/) 0,003 

(6) 0.007 (g) 0.003 

(c) 0.007 (h) 0.007 

(d) 0.003 (i) 0.023 

(e) 0.017 



MATHEMATICAL OPERATIONS 3 

The mean deviation, or average of these nine values, is 0.0092 and 
represents the amount by which an average single independent 
reading differs from the most probable value; it is therefore a 
measure of the reliability of a single observation. 

It is more important, however, to know the reliability of the 
mean than that of a single observation. It can be shown that 
the reliability of a mean or average value is numerically equal to 
the average deviation of a single observation divided by the square 
root of the number of observations taken. In the above, the 
average deviation of the mean is 0.0092/V9 = 0.0031, and the value 
for the reading may be expressed as 43.753 0.0031. (It is 
customary to use only two significant figures in all deviation meas- 
ures.) When several such measurements are involved in a com- 
putation, it is possible to calculate from the deviation measure 
of each measurement the deviation measure or precision measure 
of the final result and thus obtain a numerical measure of the 
probable reliability of that result. For methods of such calcula- 
tion the student is referred to Goodwin's Precision of Measure- 
ments. 

3. Significant Figures as a Means of Expressing Reliability. 
In most chemical analyses relatively few independent readings or 
determinations are made, so that numerical precision measures 
are not often used. In such cases the reliability or precision of a 
numerical value is best indicated by the number of significant 
figures used in expressing that value. It is true that this method 
of expression gives only an approximate idea of the reliability of 
a result, but the importance of the retention of the proper number 
of significant figures in analytical data cannot be overemphasized. 
A numerical result expressed by fewer or more significant figures 
than are warranted by the various factors involved may give to 
an observer an impression nearly as erroneous as would be given 
by a result which is inaccurate. 

4. Rules Governing the Use of Significant Figures in Chemical 
Computations. The following definitions and rules are suggested 
by those given in Goodwin's Precision of Measurements: 

A number is an expression of quantity. 

A figure, or digit, is any one of the characters 0, 1,2, 3, 4, 5, 6, 

7, 8, 9, which, alone or in combination, serve to express numbers. 

A significant figure is a digit which denotes the amount of the 



4 CALCULATIONS OF ANALYTICAL CHEMISTRY 

quantity in the place in which it stands. In the case of the 
number 243, the figures signify that there are two hundreds, four 
tens, and three units and are therefore all significant. The char- 
acter is used in two ways. It may be used as a significant 
figure, or it may be used merely to locate the decimal point. It 
is a significant figure when it indicates that the quantity in the 
place in which it stands is known to be nearer zero than to any 
other value. Thus, the weight of a crucible may be found to be 
10.603 grams, in which case all five figures, including the zeros, 
are significant. If the weight in grams of the crucible were found 
to be 10.610, meaning that the weight as measured was nearer 
10.610 than 10.609 or 10.611, both zeros would be significant. 

By analysis, the weight of the ash of a quantitative filter paper 
is found to be 0.00003 gram. Here the zeros are not significant 
but merely serve to show that the figure 3 belongs in the fifth 
place to the right of the decimal point. Any other characters 
except digits would serve the purpose as well. The same is true 
of the value 356,000 inches, when signifying the distance between 
two given points as measured by instruments which are accurate 
to three figures only. The zeros are not significant. In order to 
avoid confusion, this value should be written 3.56 X 10 5 inches. 
If the distance has been measured to the nearest 100 inches, it 
should be written 3.560 X 10 6 inches. 

Rule I. Retain as many significant figures in a result and in 
data in general as will give only one uncertain figure. (For very 
accurate work involving lengthy computations, two uncertain 
figures may sometimes be retained.) Thus, the value 25.34, rep- 
resenting the reading of an ordinary buret, contains the proper 
number of significant figures, for the digit 4 is obtained by esti- 
mating an ungraduated scale division and is doubtless uncertain. 
Another observer would perhaps give a slightly different value for 
the buret reading e.g., 25.33 or 25.35. All four figures should 
be retained. 

Rule II. In rejecting superfluous and inaccurate figures, in- 
crease by 1 the last figure retained if the following rejected figure 
is 5 or over. Thus, in rejecting the last figure of the number 
16.279, the new value becomes 16.28. 

Rule III. In adding or subtracting a number of quantities, 
extend the significant figures in each term and in the sum or dif- 



MATHEMATICAL OPERATIONS 5 

ference only to the point corresponding to that uncertain figure 
occurring farthest to the left relative to the decimal point. 

For example, the sum of the three terms 0.0121, 25.64, and 
1.05782, on the assumption that the last figure in each is uncer- 

tain, is 

0.01 

25.64 

1.06 

26.71 

Here it is seen that the second term has its first uncertain figure 
(the 4) in the hundredths place, the following figures being un- 
known. Hence, it is useless to extend the digits of the other terms 
beyond the hundredths place even though they are given to the 
ten-thousandths place in the first term and to the hundred-thou- 
sandths place in the third term. The third digit of the third term 
is increased by 1 in conformity with Rule II above. The fallacy 
of giving more than four significant figures in the sum may be 
shown by substituting x for each unknown figure. Thus, 

0.0121* 



1.05782 
26.7 Ixxx 

Rule IV. In multiplication or division, the percentage pre- 
cision of the product or quotient cannot be greater than the per- 
centage precision of the least precise factor entering into the 
computation. Hence, in computations involving multiplication 
or division, or both, retain as many significant figures in each 
factor and in the numerical result as are contained in the factor 
having the largest percentage deviation. In most cases, as many 
significant figures may be retained in each factor and in the result 
as are contained in the factor having the least member of significant 
figures. 

For example, the product of the three terms 0.0121, 25.64, 
and 1.05782, on the assumption that the last figure in each is 
uncertain, is 

0.0121 X 25.6 X 1.06 = 0.328 

for, if the first term is assumed to have a possible variation of 1 
in the last place, it has an actual deviation of 1 unit in every 



6 CALCULATIONS OF ANALYTICAL CHEMISTRY 

121 units, and its percentage deviation would be y^r X 100 = 0.8. 



Similarly, the possible percentage deviation of the second term 
would be ,, X 100 = 0.04, and that of the third term would be 



*- QO X 100 = 0.0009. The first term, having the largest per- 

lUOj/o^ 

centage deviation, therefore governs the number of significant 
figures which may be properly retained in the product, for the 
product cannot have a precision greater than 0.8 per cent. That 
is, the product may vary by 0.8 part in every hundred or by nearly 
3 parts in every 328. The last figure in the product as expressed 
with three significant figures above is therefore doubtful, and the 
proper number of significant figures has been retained. 

Rule V. Computations involving a precision not greater than 
one-fourth of l.per cent should be made with a 10-inch slide rule. 
For greater precision, logarithm tables should be used. If the 
old-style method of multiplication or division must be resorted 
to, reject all superfluous figures at each stage of the operation. 

Rule VI. In carrying out the operations of multiplication or 
division by the use of logarithms, retain as many figures in the 
mantissa of the logarithm of each factor as are properly con- 
tained in the factors themselves under Rule IV. Thus, in the 
solution of the example given under Rule IV, the logarithms of 
the factors are expressed as follows: 

log 0.0121 = 8.083 - 10 
log 25.64 = 1.409 
log 1.05782 = 0.024 

9.516 - 10 = log 0.328 

5. Conventions Regarding the Solution of Numerical Problems. 

In the calculation of numerical results from chemical data which 
have been obtained under known conditions and by known meth- 
ods, little difficulty should be experienced in forming an approxi- 
mate estimate of the reliability of the various factors and of the 
results obtained. In the case of numerical problems which are 
unaccompanied by any data to show the conditions under which 
the various measurements were made or the precision of the values 
given, the retention of the proper number of significant figures in 



MATHEMATICAL OPERATIONS 7 

the final computed results may be a matter of considerable judg- 
ment. In such cases the rules listed above are subject to modi- 
fication, but in any case the need for a certain amount of common 
sense and judgment in using them in no way detracts from their 
value. 

. In the solution of problems in this book, it may be assumed that 
the given data conform to Rule I, above. In problems containing 
such expressions as "a 2-gram sample/ 7 "a 25-ml. pipetful," or 
"a tenth-normal solution/' it may be assumed that the weight 
of the sample, the volume of the pipet, and the normality of the 
solution are known to a precision at least as great as that of the 
other factors involved in the problem. 

It should also be remembered that the atomic weights of the 
elements are known only to a limited number of significant figures 
and, in the absence of further data, it may be assumed that the 
values ordinarily given in atomic-weight tables conform to Rule I 
above, in that the last figure in each is doubtful. It follows, there- 
fore, that the same attention should be paid to the precision of 
the atomic and molecular weights involved in computations as to 
that of any other data. 

It often happens that independent calculations from given data 
give results which disagree by only one qr two units in the last 
significant figure retained. This is usually due to the fact that 
figures have been rejected at different stages of the operations 
involved; but this is usually of no importance, since, when properly 
expressed, the last significant figure in the result is doubtful 
anyway. 

Analytical determinations are usually done in duplicate. In 
most of the problems in this book, however, data apparently 
covering only one determination are given. It may be assumed 
that such values represent mean values obtained from duplicate 
determinations. 

Problems 

1. How many significant figures are implied in the value 2.20 X 10"" 9 ? In 
the value 5,000.002? In the value 2.010 X 10 5 ? 

Ans. Three. Seven. Four. 

2. Calculate the molecular weight of OsCl 4 to as high a degree of precision 
as is warranted by the atomic weights involved. 

Ans. 332.0. 



8 CALCULATIONS OF ANALYTICAL CHEMISTRY 

3. Express the velocity of light, 186,000 miles per second, in such a way as 
to indicate that it has been measured to the nearest 100 miles per second. 

Ans. 1.860 X 10 6 miles per second. 

4. Samples were sent to seven different chemists to be analyzed for per- 
centage of protein. The values reported were 43.18, 42.96, 42.88, 43.21, 
43.01, 43.10, 43.08. What is the mean value, the average deviation of a single 
value from the mean, and the deviation of the mean? If the correct percentage 
is 43.15, what is the relative error of the mean in parts per thousand? 

Ans. 43.060,0.094,0.036. 2.1. 

6. An ore actually contains 33.79 per cent Fe 2 O 8 . Duplicate determinations 
give 33.80 and 34.02 per cent, and the mean of these is reported. By how many 
parts per thousand do the duplicate results differ from each other? What is 
the mean value? What is the absolute error? What is the relative error in 
parts per thousand? 

Ans. 6.5. 33.91 per cent. 0.12 per cent. 3.5. 

6. Two analysts, working independently, analyze a sample of steel and re- 
port the following results: 

ANALYST A: ANALYST B: 

Sulfur = 0.042 per cent Sulfur = 0.04199 per cent 

0.041 per cent 0.04101 per cent 

By how many parts per thousand do the check values agree in each case? 
Each man uses a 3.5-gram sample weighed to the nearest tenth of a gram. Is 
analyst B justified in his report? Do his figures necessarily indicate greater 
ability as an analyst? 

Ans. 24 parts, 24 parts. No. No. 

7. It is necessary to solve the following: 

(1.276 X 0.00047) + (1.7 X 10~ 4 ) - (0.0021764 X 0.0121) 
each term being uncertain in the last significant figure. Should you use 
arithmetic, logarithms, or a slide rule in the multiplications? What is the 
final answer? 

Ans. Slide rule. 7.5 X 10~ 4 . 

8. A value which has been found by duplicate analyses to be 0.1129 and 
0.1133, respectively, is to be multiplied by 1.36 ml. as measured by an ordinary 
buret, and the product is to be subtracted from the value 0.93742 which has 
been very accurately measured. Express the result by the proper number 
of significant figures. 

Ans. 0.784. 

9. If in the analysis of a given substance a variation of 0.30 per cent is 
allowable, to how many milligrams should a 10-gram sample be weighed? 

10. How many significant figures are implied in the value 16 X 10 3 ? In the 
value 16.00 X 10 3 ? In the value 1.60 X 10~ 2 ? 



MATHEMATICAL OPERATIONS 9 

11. In the following multiplication the last figure in each of the three 
factors is uncertain. How many figures in the product as given should be 
rejected as superfluous? Express the product in such a way as to indicate the 
correct number of significant figures. 

2.0000 X 0.30 X 500 = 300.00 

12. Calculate the molecular weight of Hf (NO 8 )4 to as high a degree of pre- 
cision as is warranted by the atomic weights involved. 

13. A book on astronomy gives the polar diameter of the earth as 7,900.0 
miles. To what precision of measurement does this number imply? If the 
measurement had been made only to the nearest 10 miles, how should the value 
be expressed to indicate this fact? 

14. Assuming each term to be uncertain in the last figure given, solve the 
following and express the answer to the correct number of significant figures: 

(1.586 ^ 29.10) + [162.22(3.221 X 1Q- 4 )] - 0.00018 

16. A sample of pure anhydrous BaCl 2 containing 65.97 per cent Ba is given 
for analysis. One analyst obtains 65.68, 65.79, and 65.99 for triplicate determi- 
nations and reports the mean value. By how many parts per thousand does 
each result differ from the mean? What is the absolute error of the mean, and 
what is the relative error (parts per thousand) of the mean? 

16. The percentage of carbon in a sample of steel is found to be 0.42 per 
cent. The calculations involve only multiplication and division. To how 
many decimal places would you weigh out a 1-gram sample in order to duplicate 
the result? 

17. A sample of limonite was analyzed by 12 students at different times 
during the college year. The values obtained for the percentage of iron 
were: 34.62, 34.42, 34.60, 34.48, 33.71, 34.50, 34.50, 34.22, 34.41, 35.00, 34.65, 
34.44. What is the mean value, the mean deviation of a single result, and the 
deviation of the mean? If the correct percentage is 34.75 what is the absolute 
error of the mean and what is its relative error in parts per thousand? 

18. A sample of material was sent to two chemists. Each used the same 
method and reported the results of four analyses, as follows: 

CHEMIST A CHEMIST B 

30.15 30.251 

30.15 30.007 
30.14 30.101 

30.16 30.241 

Calculate in each case the mean value and its deviation measure. Other 
conditions being equal, which mean value is the more reliable? 

6. Rules Governing the Use of Logarithms. In calculations of 
quantitative analysis involving multiplication and division where 
four-significant-figure accuracy is required, four-place logarithms 



10 CALCULATIONS OF ANALYTICAL CHEMISTRY 

should be used; in calculations where two- or three-significant- 
figure accuracy is sufficient, a slide rule should be used. Grammar- 
school methods of multiplication and long division should not be 
employed. 

Although the theory and use of logarithms are ordinarily covered 
in preparatory and high schools, the following outline is given as 
a review of the essential points in this phase of mathematics. 

1. The logarithm of a number is the exponent of the power to 
which some fixed number, called the base, must be raised to equal 
the given number. Thus, suppose 

a x = n 

then x is the logarithm of n to the base a and may be written 

x = log a n 

2. The base in the common system of logarithms is 10, and the 
term log, without subscript, is commonly used to denote a log- 
arithm in this system. Hence, 

10 - 1, log 1 =0 

10 1 = 10, log 10 - 1 

10 2 - 100, log 100 = 2 

10 3 = 1000, log 1000 - 3 
lO-^O.l, log 0.1 = -1 
10~ 2 = 0.01, log 0.01 = -2 
10~ 3 = 0.001, log 0.001 = -3 

etc. 
It is evident that the logarithms of all numbers between 

1 and 10 will be plus a fraction 

10 and 100 will be 1 plus a fraction 

100 and 1000 will be 2 plus a fraction 

1 and 0.1 will be 1 plus a fraction 

0.1 and 0.01 will be 2 plus a fraction 

etc. 

3. If a number is not an exact power of 10, its common logarithm 
can be expressed only approximately as a number with a continu- 
ing decimal fraction. Thus, 

36= ioi--- 
or 

log 36= 1.5503---- 



MATHEMATICAL OPERATIONS 11 

The integral part of the logarithm is called the characteristic; 
the decimal part is called the mantissa. In the case just cited, 
the characteristic is 1; the mantissa is .5563. Only the mantissa 
of a logarithm is given in a table of logarithms (see next section) ; 
the characteristic is found by means of the next two rules. 

4. The characteristic of the logarithm of a number greater than 
1 is 1 less than the number of digits to the left of the decimal point. 
For example, the characteristic of log 786.5 is 2; the characteristic 
of log 7.865 is 0. 

5. The characteristic of the logarithm of a decimal number be- 
tween and 1 is negative and is equal in numerical value to the 
number of the place occupied by the first significant figure of the 
decimal. For example, the characteristic of log 0.007865 is 
-3. 

6. The mantissa of a logarithm is always positive; the charac- 
teristic may be either positive or negative. For example, 

log 36.55 = +1 + .5629 = 1.5629 
log 0.08431 = -2 + .9259 

This last logarithm is more conventionally written as 2.9259 
with the understanding that only the 2 is negative. Another 
common method of expressing this logarithm is 8.9259 10. 

7. The mantissas of the common logarithms of numbers having 
the same sequence of figures are equal. For example, 

log 2.383 = 0.3772 
log 23.83 - 1.3772 
log 0.002383 = 3.3772 (or 7.3772 - 10) 

8. The cologarithm of a number is the logarithm of the recip- 
rocal of that number. It is found by subtracting the logarithm 
of the number from zero. For example, 

log 7. 130 = 0.8531 
colog 7.130 - 0.0000 - 0.8531 

= 1.1469 
or 

10.0000 - 10 
0.8531 
colog 7.130 = 9.1469 - 10 



12 CALCULATIONS OF ANALYTICAL CHEMISTRY 

9. The antilogarithm of A is the number that has A for a 
logarithm. For example, 

log 7.130 =0.8531 
antilog 0.8531 = 7.130 

10. The logarithm of a product is equal to the sum of the 
logarithms of its factors. For example, 

log (7.180 X 586.3) - log 7.180 + log 586.3 
= 0.8531 + 2.7681 
= 3.6212 

11. The logarithm of a fraction is equal to the 'logarithm of 
the numerator minus the logarithm of the denominator; it is also 
equal to the logarithm of the numerator plus the cologarithm of 
the denominator. For example, 

7 180 
log ~ = log 7.180 - log 586.3 



= 0.8531 - 2.7681 

- 2.0850 (or 8.0850 - 10) 
or 

7 ISO 
log = log 7.180 + colog 586.3 



= 0.8531 + 3.2319 (or 7.2319 - 10) 
= 2.0850 (or 8.0850 - 10) 

The use of cologarithms is particularly advantageous when the 
multiplication and division of several factors are involved in the 
same mathematical process. This is shown in the example at 
the end of this section. 

12. The logarithm of any power of a quantity is equal to the 
logarithm of the quantity multiplied by the exponent of the power. 
For example, 

log 71. 80 3 = 3 X log 71.80 
= 3 X 1.8531 
- 5.5593 

13. The logarithm of any root of a quantity is equal to the 
logarithm of the quantity divided by the index of the root. For 
example, 

log ^002 = Y 2 X log 5.002 
= y 2 X 0.6992 
= 0.3496 



MATHEMATICAL OPERATIONS 13 

7. Method of Using Logarithm Tables. The precision of or- 
dinary chemical analytical work is seldom great enough to permit 
the retention of more than four significant figures in the numerical 
data obtained and in the calculations made from such data. Hence 
a four-place logarithm table such as is given in the back of this 
book is entirely adequate. 

To use the logarithm table in finding a mantissa proceed as 
follows: First find the first two digits of the number in the column 
headed " natural numbers," then go to the right until the column 
is reached which has the third digit of the number as a heading. 
To the number thus found add the number which is in the same 
horizontal line at the right-hand side of the table and in the column 
of proportional parts headed by the fourth significant figure of 
the number. Thus the number representing the mantissa of log 
236.8 is 3729 + 15 = 3744, and the logarithm is 2.3744. 

Antilogarithms may be looked up in the antilogarithm table in 
the same way. Only the mantissa is used in looking up the num- 
ber; the characteristic is used merely to locate the decimal point. 
Thus the sequence of digits in the number having a logarithm of 
1.8815 is 7603 + 9 = 7612, and the actual number is 76.12 as de- 
termined by the given characteristic of the logarithm. 

In actual calculations from analytical data the essential purpose 
of the characteristic in a logarithm is to locate the position of the 
decimal point in the final numerical value obtained. Since in most 
cases a very rough mental calculation is all that is needed to 
establish the position of the decimal point, the use of character- 
istics can be dispensed with. The retention of characteristics is, 
however, helpful in serving as a check on the other method. 

Calculations of quantitative chemical analysis in which loga- 
rithms are of value seldom involve operations other than those of 
multiplication and division. 

v ^ i i * u i -xi 9.827 X 50.62 

LxAMPLE.-Calculate by logarithms: 0tQ05164 x 136>6Q - 

SOLUTION: 

Method A (without using cologarithms) 

log 9.827 = 0.9924 

log 50.62 = 1.7044 

Sum = 2.6968 



14 CALCULATIONS OF ANALYTICAL CHEMISTRY 

log 0.005164 = 3.7129 or 7.7129 - 10 
log 136.59 = 2.1354 2.1354 

Sum = T.8483 9.8483 - 10 

log numerator = 2.6968 or 12.6968 - 10 
log denominator = 1.8483 9.8483 - 10 

Difference = 2.8485 2.8485 

antilog = 705.5. Ans. 

Method B (using cologarithms) 

log 9.827 = 0.9924 or 0.9924 

log 50.62 = 1.7044 1.7044 

colog 0.005164 = 2.2871 2.2871 

colog 136.59 = 3.8646 7.8646 - 10 

Sum = 2.8485 12.6485 - 10 

antilog = 705.5. Ans. 

As previously mentioned, much time is saved by omitting all 
characteristics in the solution of the above problem and merely 
writing down the mantissas of each logarithm or cologarithm. 
The location of the decimal point is then determined by a simple 
mental calculation on the original expression. Thus, inspection 
shows that the two factors in the numerator of the above expres- 
sion give a result approximating 500 and that the factors in the 
denominator give a result approximating 0.7. The answer must 
therefore be in the neighborhood of 700, which establishes the 
position of the decimal point. 

8. Use of the Slide Rule. The slide rule is essentially a loga- 
rithm table, mechanically applied. On the scales used for multipli- 
cation and division the numbers are stamped on the rule in positions 
proportionate to their logarithms. Multiplication by means of 
the rule is merely a mechanical addition of two logarithms; divi- 
sion is a mechanical subtraction of two logarithms. Manuals cov- 
ering the proper use of a slide rule are readily obtainable and are 
usually provided by the manufacturer of the rule. 

The student of quantitative analysis should be proficient in the 
use of a slide rule, particularly in the processes of multiplication 
and division. The slide rule saves a great deal of time in making 
minor calculations and is an excellent means of checking calcula- 
tions made by logarithms. Although the precision of the ordinary 



MATHEMATICAL OPERATIONS 15 

10-inch slide rule is limited to three significant figures, it is sug- 
gested that slide-rule accuracy be permitted in solving quiz prob- 
lems and home problems, even though the data given may theo- 
retically require four-significant-figure accuracy. The purpose of 
the problems is more to make sure that the methods of calculation 
are understood than to give practice in fundamental mathematical 
operations. 

Most laboratory calculations, however, require four-significant- 
figure accuracy, and four-place logarithms are necessary. 

Problems 

19. Using four-place logarithms determine the following: (a) log 387.6, 
(6) log 0.0009289, (c) colog 52.61, (d) colog 0.06003, (e) antilog 2.4474, (/) anti- 
log 4.1733, (g) aritilog 7.2068 - 10. 

Ans. (a) 2.5884, (b) 4.9679 or 6.9679 - 10, (c) 2.2789 or 8.2789 - 10, 
(d) 1.2216, (e) 280.2, (/) 0.0001490, (g) 0.001610. 

20. Using four-place logarithms calculate the following: (a) 226.3 X 
0.00002591, (6) 0.05811 + 64.53, (c) fourth power of 0.3382, (d) cube root 
of 0.09508. Check these to three significant figures with a slide rule. 

Ans. (a) 0.005864, (6) 0.0009005, (c) 0.01308, (d) 0.4564. 

21. Using four-place logarithms find the value of the following. Locate 
the position of the decimal point by mental arithmetic and also by the proper 
use of characteristics. Also check the answer to three significant figures with 

a slide rule. 

0.0046191 X 287.7 
51.42 X 0.84428 
Ans. 0.03061. 



22. Using four-place logarithms determine the following: (a) log 67.84, 
(6) log 0.005903, (c) colog 0.9566, (d) colog 718.1, (e) antilog 3.6482, (/) anti- 
log 2.0696, (g) antilog 6.0088 - 10. 

23. Using four-place logarithms calculate the 'following: (a) 33.81 X 
0.0009915, (6) 0.1869 -5- 362.4, (c) cube of 0.09279, (d) square root of 
0.5546. Check these to three significant figures with a slide rule. 

24. Using four-place logarithms find the numerical value of the following 
expression. Locate the position of the decimal point by the proper use of 
characteristics and check by mental arithmetic. Also check the answer to 
three significant figures by means of a slide rule. 

5987.9 X 0.006602 
1.864 X 0.4617 X 1053.3 



CHAPTER II 
CHEMICAL EQUATIONS 

9. Purpose of Chemical Equations. When the nature and com- 
position of the initial and final products of a chemical reaction 
are known, the facts can be symbolized in the form of a chemical 
equation. When properly written, the equation indicates (1) the 
nature of the atoms and the composition of the molecules taking 
part in the reaction, (2) the relative number of atoms and mole- 
cules of the substances taking part in the reaction, (3) the pro- 
portions by weight of the interacting and resulting substances, 
and (4) the proportions by volume of all gases involved. These 
four principles applied to reactions which go to completion serve 
as the foundation of quantitative chemical analysis. Before the 
calculation of a chemical analysis can be made, it is important to 
understand the chemistry involved and to be able to express the 
reactions in the form of balanced equations. 

10. Types of Chemical Equations. The determination of the 
nature of the products formed by a given reaction involves a 
knowledge of general chemistry which, it is assumed, has already 
been acquired from previous study, but the ability to write and 
balance equations correctly and quickly is acquired only by con- 
siderable practice. The following discussion is given to help the 
student attain this proficiency, especially in regard to equations 
involving oxidation and reduction, which usually give the most 
trouble to the beginner. 

With equations expressing the reactions of (1) combination, 
(2) decomposition, and (3) metathesis, it is seldom that much 
difficulty is experienced in bringing about equality between the 
atoms and molecules of the reacting substances and those of the 
products, for little more is involved than purely mechanical ad- 
justment of the terms and an elementary knowledge of valence. 
As examples of the above types of chemical change in the order 
given, the following equations may be cited : 



1ft 



CHEMICAL EQUATIONS 17 



(1) 

(2) 2HgO -> 2Hg + O 2 

(3) FeCl 3 + 3NH 4 OH -* Fe(OH) 3 + 3NH 4 C1 

Equations expressing reactions of oxidation and reduction, al- 
though usually somewhat more complicated, offer little additional 
difficulty, provided that the principles underlying these types of 
chemical change are thoroughly understood. 

The above equations are molecular equations. For reactions 
taking place in aqueous solution (such as the third case above) 
equations are usually better written in the ionic form. To do so 
correctly requires a knowledge of the relative degrees of ionization 
of solutes and the correct application of a few simple rules. 

11. Ionization of Acids, Bases, and Salts. Although the theory 
of ionization should be familiar to the student from his previous 
study of general chemistry, the following facts should be kept in 
mind because they are particularly important in connection with 
writing equations: 

" Strong " acids include such familiar acids as HC1, HBr, HI, 
H 2 S0 4 , HN0 3 , HC10 3 , HBr0 3 , HIO 3 , HC1O 4 , and HMnO 4 . These 
acids in solution are 100 per cent ionized, although at ordinary 
concentrations inter-ionic effects may give conductivities corre- 
sponding to an apparent degree of ionization a little less than 
100 per cent. In ionic equations (see below) strong acids are 
written in the form of ions. 

"Strong" bases include NaOH, KOH, Ba(OH) 2 , Sr(OH) 2 , and 
Ca(OH) 2 . These bases in solution are 100 per cent ionized and 
in ionic equations are written as ions. 

Salts, with very few exceptions, are completely dissociated into 
simple ions in solution, and in ionic equations are written as ions. 
Two common exceptions are lead acetate and mercuric chloride. 

Many acids and bases are ionized in solution to only a slight 
degree at ordinary concentrations. Table IX in the Appendix 
lists most of such acids and bases ordinarily encountered in ana- 
lytical chemistry, and the student should familiarize himself with 
the names of these substances and have at least a general idea ol 
the magnitude of the degree of ionization in the case of the mon 
common ones. 

Certain acids contain more than one hydrogen replaceable bj 



18 CALCULATIONS OF ANALYTICAL CHEMISTRY 

a metal (polybasic acids). It will be noted that these acids ionize 
in steps, and the degree of ionization of the first hydrogen is in- 
variably greater than that of the others. Phosphoric acid, for 
example, is about 30 per cent ionized in tenth-molar solution to 
give H + and H 2 PO 4 ~~ ions, but the concentration of HPO 4 = ions 
is much less, and that of PQ 4 S ions is very small. Sulfuric acid is 
100 per cent ionized into H+ and HSO 4 ~ ions, but the bisulfate 
ion is only moderately ionized further to give H + ions and SCV* 
ions. 

12. Ionic Equations Not Involving Oxidation. Most of the re- 
actions of analytical chemistry are reactions between ions in solu- 
tion. For this reason, although the molecular type of equation 
is serviceable as a basis for quantitative analytical calculations, 
the so-called ionic equation is usually easier to write and is gen- 
erally better. 

In writing ionic equations, the following basic rules should be 
observed : 

1. Include in the equation only those constituents actually taking 
part in the chemical reaction. 

EXAMPLE I. The addition of a solution of sodium hydroxide 
to a solution of ferric nitrate results in a precipitation of ferric 
hydroxide. The ionic equation is as follows: 

Fe+++ + 30H- -> Fe(OH) 3 * 

The sodium ions from the sodium hydroxide and the nitrate ions 
from the ferric nitrate do not enter into the reaction and hence 
are not represented in the equation. 

2. In cases where a reac,tant or product exists in equilibrium with 
its constituent ions, express in the equation that form present in 
greatest amount. 

It follows that weak acids, weak bases, and the slightly ionized 
salts should be written in the molecular form. Substances of this 
type most often encountered in analytical chemistry are the 
following: H 2 O, HC 2 H 3 O 2 , NH 4 OH, H 2 S, H 2 C0 3 , HNO 2 , HF, 
Pb(C 2 H 3 2 ) 2 , HgCl 2 , H 3 P0 4 , H 2 C 2 O 4 , and H 2 SO 3 (see Table IX, 
Appendix). The last three of these are borderline cases since they 

* It is desirable to underline formulas of precipitates. The use of down- 
ward-pointing arrows is equally satisfactory. If desired, formulas of gases 
may be overlined or denoted by upward-pointing arrows. 



CHEMICAL EQUATIONS 19 

are ionized to a moderate degree to give hydrogen ions and 
H 2 PO 4 ~, HC 2 O 4 ~, and HSO 3 ~ ions, respectively. The salts lead 
acetate and mercuric chloride may be dissociated somewhat into 
complex ions [e.gr., Pb^HaC^)" 1 " in the former case] but are rela- 
tively little ionized to give the metal ions. They are therefore 
usually written in the molecular form. 

EXAMPLE II. The addition of an aqueous solution of ammo- 
nium hydroxide to a solution of ferric nitrate results in a pre- 
cipitation of ferric hydroxide. The ionic equation is as follows: 

Fe+++ + 3NH 4 OH -> Fc(OH) 3 + 3NH 4 + 

In this case, although ammonium hydroxide is ionized into am- 
monium ions and hydroxyl ions, the ionization is comparatively 
slight and only the undissociated ammonium hydroxide molecules 
are expressed in the equation. 1 

EXAMPLE III. The addition of a solution of hydrogen sulfide 
to an acid solution of copper sulfate gives a precipitate of copper 

sulfide: 

Cu++ + H 2 S -> CuS + 2H ' 

The fact that the original solution is acid docs not require that 
hydrogen ions be on the left-hand side of the equation. The equa- 
tion merely indicates that the solution becomes more acid. 

EXAMPLE IV. When a solution containing lead nitrate is treated 
with sulfuric acid, a white precipitate of lead sulfate is obtained. 
This precipitate dissolves in a solution of ammonium acetate, and 
the addition of a solution of potassium chromate then causes a 
yellow precipitate to appear. The ionic equations for these re- 
actions are 

Pb++ + HS0 4 ~ -> PbS0 4 + H+ 

PbSO 4 + 2C 2 H 3 2 - - Pb(C 2 H 3 O 2 ) 2 + S0 4 - 

Pb(C 2 H 3 O 2 ) 2 + CrO 4 - -* PbCr0 4 + 2C 2 H 3 O 2 - 

1 As a matter of fact, it is not entirely certain that an appreciable concen- 
tration of NH 4 OH exists at all. The equilibrium existing in an aqueous solu- 
tion of ammonia is more generally and better expressed as follows: 

NH 3 + H 2 O <=> [NH 4 OH (?)] ^ NH 4 + 



Here again, the concentration of OH"~ is relatively low, and the equation for 
the above reaction can therefore be written 

3NH 3 + 3H 2 O -> Fe(OH) 8 + 3NH 4 + 



20 CALCULATIONS OF ANALYTICAL CHEMISTRY 

EXAMPLE V. Silver chloride dissolves in an aqueous solution 
of ammonia. The equation is written as follows (see Example II 
and footnote above) : 

AgCl + 2NH 4 OH -> Ag(NH 3 ) 2 + + 01- + H 2 
or 

AgCl + 2NH 3 - Ag(NH 3 ) 2 + + 01- 

The silver ammino ion, like most complex ions, is only very slightly 
dissociated into its constituents: Ag(NH 3 ) 2 + > Ag+ + 2NH 3 . 

EXAMPLE VT. A nitric acid solution of ammonium molybdate 
((NH 4 ) 2 MoO 4 ] added to a solution of phosphoric acid results in 
the precipitation of ammonium phosphomolybdate. 

12MoOr + H 3 P0 4 + 3NH 4 + + 21H+ -> 

(NH 4 ) 3 P0 4 .12Mo0 3 + 12H 2 O 

Note here that for every 12 molybdate ions only 3 of the corre- 
sponding 24 ammonium ions present enter into the reaction. The 
nitrate ions of course take no part in the reaction. 

13. Oxidation Number. Although the term "valence" usually 
refers to the degree of combining power of an atom or radical, it 
is likely to be applied somewhat differently in the various branches 
of chemistry. For this reason, in inorganic chemistry the term 
"oxidation number" is to be preferred in expressing state of 
oxidation. 

It is assumed that the student is already familiar with the 
general aspects of the periodic table and with the combining power 
of the elements he has thus far studied. It will be recalled that 
(1) the oxidation number of all free elements is zero, (2) the oxida- 
tion number of hydrogen in its compounds is +1 (except in the 
case of the relatively rare metallic hydrides), (3) the oxidation 
number of sodium and potassium in their compounds is +1, and 
(4) the oxidation number of oxygen in its compounds is 2 (with 
few exceptions). 

Since the algebraic sum of the oxidation numbers of the elements 
of a given compound is zero, the oxidation number of any element 
in a compound can usually be readily calculated from those of the 
other elements making up the compound. Thus, the oxidation 
number of Cl in HC1O 3 is +5, since +1 + 5 + [3 X (-2)] - 0. In 
this case the oxidation number of the C10 3 radical is 1, since it is 
combined with the +1 hydrogen. The oxidation number of S 2 in 



CHEMICAL EQUATIONS 21 

is +12 since Na 2 = +2 and 7 oxygen atoms = 14. Each 
sulfur atom therefore has an oxidation number of +6. 

The oxidation number of an ion is the same as the charge it 
bears. Thus, the oxidation number of the nitrate ion (N0 3 ~) is 
1, that of the sulfate ion (SC^) is -2, and that of the phos- 
phate ion (P0 4 S ) is -3. 

A few cases may give trouble. Thus, in the compound HCNO 
the sum of the oxidation numbers of the carbon and nitrogen atom 
is obviously +1, but this would be true if C = +4 and N = 3,. 
or C = +3 and N = -2, or C = +2 and N = -1, etc. However, 
since the oxidation number of carbon is so often +4 (e.g., GO 2 > 
and that of nitrogen is so often -3 (e.g., NH 3 ), these would be- 
the most likely oxidation numbers to take. 

A compound like Fe 3 O 4 shows an apparent fractional oxidation 
number for the metal constituent, in this case 2%. Actually two 
of the iron atoms have an oxidation number of +3, and one iron 
atom has an oxidation number of +2. This is called a mixed 
oxide (FeO.Fe 2 3 ). A similar case is the salt Na 2 S 4 O 6 ; the average, 
oxidation number of each sulfur atom is 2^. 

In so-called per-oxy acids and salts of these acids, one (,or more) 
of the oxygen atoms has an oxidation number of zero. For ex- 
ample, in hydrogen peroxide, H 2 2 , the oxidation number of one 
oxygen atom is 2; that of the other is 0. Sulfur forms analogous 
per-sulfur acids. 

14. Ionic Oxidation and Reduction Equations. In the case of 
oxidation-reduction equations the two rules given in Sec. 12 should 
also be observed. It will be found convenient in most cases to 
write equations systematically according to the following steps: 

a. Write the formula of the oxidizing agent and of the reducing 
agent on the left-hand side of the equation. These should conform 
to Rules 1 and 2. 

b. Write the formulas of the resulting principal products on the 
right-hand side of the equation. These should likewise conform to 
Rules 1 and 2. 

c. Under the formula of the oxidizing substance, write the number 
expressing the total change in oxidation number of all of its con- 
stituent elements. Under the formula of the reducing substance, write 
the number expressing the total change in oxidation number of its 
constituent elements. 



22 CALCULATIONS OF ANALYTICAL CHEMISTRY 

d. Use the number under the formula of the oxidizing agent in the 
equation as the coefficient for the reducing substance; use the number 
under the formula of the reducing agent in the equation as the co- 
efficient for the oxidizing substance. 

e. Insert coefficients for the principal products to conform to the 
preceding step. 

f. If possible, divide all the coefficients by the greatest common 
divisor, or, if necessary, clear of fractions by multiplying all the 
coefficients by the necessary factor. 

g. If the reaction takes place in acid solution, introduce the for- 
mulas H^O and //+ in amounts necessary to balance the atoms of 
oxygen and hydrogen on the two sides of the chemical equation. If 
the reaction takes place in basic solution, introduce the formulas 
HtO and OH" in amounts necessary to balance the atoms of oxygen 
and hydrogen. 

h. Check the equation by determining the total net ionic charge on 
each of the two sides of the equation. They should be the same. 

EXAMPLE I. When a solution of chlorine water is added to 
a sulfuric acid solution of ferrous sulfate, the iron is oxidized. 
The step-by-step formulation of the equation for this reaction is 
as follows: 

STEP RESULT 

a, 6 Fe ++ + C1 2 -> Ke +++ + Cl~ 
c Fe++ + C1 2 -> Fe^ + + Gl~ 

1 2 

d, e 2FeV + C1 2 -> 2Fe 4++ + 2C1~ 
/ None 

g None 

h 4+ = 4 + 

EXAMPLE II. When a dilute nitric acid solution of stannous 
chloride is treated with a solution of potassium dichr ornate, tin 
is oxidized (from 2 to 4) and chromium is reduced (from 6 to 3). 
Neglecting the partial formation of complex ions (e.g. Slide") the 
development of the equation is as follows: 

STEP RESULT 

a, 6 Sn ++ + Cr 2 Or -> Sn++++ 

c Sn+ + + Cr 2 O 7 - -> 811++++ + Cr 
2 34-3 

rf, e 6Sn++ + 2Cr,O7- -* 6811++++ + 

/ 3Sn+ + + Cr 2 O 7 ~ -> 3Sn +++ + + 

g 3Sn++ + Cr 2 O 7 ~ + 1411+ -> 3Sn++++ + 2Cr+ ++ + 7H 2 O 

h 18+ = 18+ 



CHEMICAL EQUATIONS 23 

Note that in writing this equation in the molecular form one 
would be at a loss to express the products correctly. The ques- 
tion would arise whether to write stannic chloride and chromic 
nitrate or stannic nitrate and chromic chloride. As a matter of 
fact, none of these is formed since the salts are completely ionized 
in dilute solution. 

EXAMPLE III. When hydrogen sulfide is bubbled into a dilute 
sulfuric acid solution of potassium permanganate, the latter is 
reduced (to manganous salt) and a white precipitate of free sulfur 
is obtained. 

STEP RESULT 

a, b MnOr + H 2 S -> Mn++ + S 

c MnOr + H 2 S -> Mn+ + + S 

5 2 

d, e 2MnOr + 5HiS -> 2Mn++ + 5S 

/ None 

g 2MnOr + 5H 2 S + 6H+ - 2Mn++ + 58 + 8II 2 O 

h 4+ = 4+ 

EXAMPLE TV. In the presence of sulfuric acid an excess of 
potassium permanganate solution will oxidize a chromic salt to 
clichromatc. 

RESULT 

t- Mn++ 



STEP 






RESULT 


a, b 


Cr+ 4+ - 


f MnOr 


-> Cr 2 () 7 - 


c 


( i r +++ 


f MnOr 


- Cr 2 7 - 




3 


5 




d, c 


5Cr ^ + - 


f SMnOr 


> Ol/f yf ( 


i 


10Cr ++f - - 


f GMnOr 


> 5( ^l^Oy' 


<J 


10Cr++ + - 


f 6Mn()r 


+ 11FI 2 O 


h 




24+ 


= 24+ 



3Mn + - 
6Mn f + 

+ 6Mn ++ + 22TI- 1 



EXAMPLE V. When metallic aluminum is added to a solution 
of a nitrate in caustic alkali, the latter is reduced and ammonia, 
gas is evolved. 

STEP RESULT 

a, 6 Al + N0 3 - - AlOr + NH 3 

c Al +N0 3 - -A10- +NH, 

3 8 

d, e 8A1 + 3NOr - 8A1O 2 ~ + 3NH 3 
f None 

g 8A1 + 3NO 3 - + 5OH- + 2H 2 O - SAlOr + 3NH 3 
h 8- = 8- 



24 CALCULATIONS OF ANALYTICAL CHEMISTRY 

EXAMPLE VI. Solid cuprous sulfide is oxidized by hot con- 
centrated nitric acid forming a cupric salt, sulfate, and N0 2 gas* 

STEP RESULT 

*. b CutS + NO 3 - - Cu++ + SO 4 - + NO 2 
c CujS + NO 3 - - Cu++ + SO 4 ~ + NO 2 

1+1+8 1 

d, e CuaS + 10NO 3 - -* 2Cu ++ + SOr + 10NO 2 

None 
g CuzS + 10NO 3 - + 12H+ -* 2Cu++ + SO 4 - + 10NO 2 + 6H 2 O 

h 2+ = 2+ 

Problems 

26. What is the oxidation number of each of the elements (other than 
hydrogen and oxygen) in each of the following: (a) N 2 O 3 ; (6) SbS 3 = ; (c) H 4 P 2 O S ; 
(d) K 2 Pt(N0 2 ) 4 ; (e) S 8 ; (/) Co(NH 3 ),+++; (g) Cu 3 [Fe(CN) 6 ] 2 ; (h) NaCHO 2 ? 

Ans. (a) +3; (b) +3, -2; (c) +3; (d) +1, +2, +3; (e) 0; (/) +3, -3; 
(O) +2, +3, +2, -3; (A) +1, +2. 

26. What is the oxidation number of each of the elements (other than 
hydrogen and oxygen) in each of the following: (a) MnO 2 ; (6) A1 2 (SO 4 ) 3 ; 
(c) NaCu(CN) 2 ; (d) (VO) 3 (PO 4 ) 2 ; (e) Fe(ClO 3 ) 3 ; (/) HAsOr; (?) CdS 2 O 6 .6H 2 O; 
(h) (UO 2 )(C1O 4 ) 2 .4H 2 O? 

Ans. (a) +4; (6) +3, +6; (c) +1, +1, +2, -3; (d) +4, +5; (e) +3, 
+5; (/) +5; (g) +2, +5; (h) +6, +7. 



27. State the oxidation number of each element in the following: (a) N 2 ; 
(6) N 2 0; (c) H 2 Se; (d) Mn 2 O 7 ; (e) HCN; (/) K 4 Fe(CN) 6 ; (g) Na 2 Cr 2 O 7 ; 
<A) (NH 4 ) 2 SO 4 . 

28. State the oxidation number of each element in the following: 
(a) K 2 Mn0 4 ; (6) IIC1O 4 ; (c) Mg 2 P 2 O 7 ; (d) Pb 3 O 4 ; (e) K 2 II 2 Sb 2 O 7 ; (/) Na 2 S 2 ; 
(g) Na 2 S 2 O.,.5H 2 O; (h) NH 2 OH; (t) IIN 3 . 

29. State the oxidation number of each element in the following: (a) A1O 2 ~; 
(6) Fe(CN) 6 B ; (c) Cu(NII 3 ) 4 ++; (d) Ag(CN) 2 -; (e) SnS 3 ~; (/) MgNH 4 AsO 4 ; 
(g) Na 2 B 4 7 ; (h) HC 2 O 4 -; (0 WF 8 -. 

30. Give the oxidation number of each element in the following: 
(a) K 2 PtCl c ; (b) (UO 2 ) 3 (PO 4 ) 2 ; (c) K 3 Co(NO 2 ); (d) SbOCl; (e) HC 2 H 3 O 2 ; 
(/) LiH; (g) Bi(OH)CO 3 ; (h) Hgl.HgNHJ; (0 Fe 4 [Fe(CN) 6 ] 3 . 

31. How do you account for the unusual average oxidation number of sulfur 
in (a) Na 2 S 4 O 6 ; (6) FeS 2 ; (c) Na 2 S 2 O 8 ? 

32. The following unbalanced equations do not involve oxidation and re- 
duction. Convert them into complete, balanced ionic equations. Introduce 
H 2 O and other constituents wherever necessary. Substances are in solution 
unless underlined. 

(a) A1C1 3 + NaOH -> NaAlO 2 

(6) Fe 2 (SO 4 ) 3 + NH 4 OH ->Fe(OH) 3 



CHEMICAL EQUATIONS 26- 

(c) CuSO + NH 4 OH - Cu(NH,) 4 SO 4 

(d) K 2 Cd(CN) 4 + H 2 S -Cd 

(e) FeCl 3 + K 4 Fe(CN) -> Fe 4 [Fe(CN) 6 l 3 

(/) H 3 PO 4 + (NH 4 ) 2 Mo0 4 + HNO 3 -> (NH 4 ),PO 4 .12MoO, 
(g) Na 3 AsS 4 + H 2 SO 4 ->As 2 S t + 1I 2 S 
(h) Na,Sb0 4 + II 2 SO 4 + H 2 S -> Sb 2 S 5 
HC1 -> U0 2 C1 2 



0) HC 2 H 3 O 2 + PbO - Pb(C 2 H 3 O 2 ) 2 

33. The following unbalanced equations do not involve oxidation and 
reduction. Introduce the necessary constituents and convert to complete, 
balanced, ionic equations. Substances are in solution unless underlined. 
(a) 



(&) Pb(OH) 2 + KOH -> KiPbOi 

(c) AgCl + NH 4 OH - Ag(NII,) 2 Cl 

(d) H 2 SnCle + H 2 S -8nS, 

(e) Sb 2 S 6 + (NH 4 ) 2 S -> (NH^jSbSi 
(/) Hg(N0 3 ) 2 + KI -> K 2 HgI 4 

to) Na 2 SnO 3 + HC1 + H 2 S -> SnS 2 

(K) UO 2 SO 4 + KOH->K 2 U 2 O 7 

() Pb(C 2 H 3 O 2 ) 2 + K 2 CrO 4 - PbCrO 4 + KC 2 H,O 2 

34. Balance the following oxidation and reduction equations: 
(a) Fe ++ + ClOr + H + - Fe +++ + Cl" + H 2 O 
(6) Cr ++ -'- + MnO 2 + H 2 -> Mn++ + CrOr + H^ 

(c) MnOr + a- + H+ -> Mn ++ + C1 2 + H 2 O 

(d) MnO 4 ~ + H 2 S + H+ -* Mn++ + S + H 2 O 

(e) I0 3 - + I- + H + -> I 2 + II 2 

(/) MnO 4 ~ + S 2 O 3 - + H+ - Mn++ + S 4 O 6 - + H 2 O 

36. Complete and balance the following ionic oxidation-reduction equations: 
(a) MnOr + I- + H+ -* Mn++ + I 2 + H 2 O 
(6) Cr 2 7 - + H 2 S + H+ -* Cr+++ + S + H 2 O 

(c) Zn + OH- - ZnO 2 - + H 2 

(d) AsO 4 - + Zn + H + -> AsH 3 + Zn++ + H 2 O 

(e) BrOr + I" + H+ -* Br~ + I 2 + H 2 O 

(/) NOr + Al + OH- + H 2 O -* NH 3 + AlOr 
to) Cr + ++ + NaA + OH- - CrOr + Na+ + II 2 O 
(K) Al+ ++ + S 2 3 - ->A1(QH) 3 + 8 + S0 2 

36. The following unbalanced oxidation-reduction equations represent 
reactions taking place in acid solution. Convert them to balanced ionic 
equations, introducing H + and H 2 O wherever necessary. 

(o) Cr 2 O 7 - + NO 2 - - Cr+++ + NO 3 - 

(6) Cr +++ + S 2 8 - - Cr 2 O 7 - + SO 4 ~ 

(c) MnOr + H 2 O 2 -* Mn++ + O 2 



26 CALCULATIONS OF ANALYTICAL CHEMISTRY 



(d) Mn++ + BiO 2 -> MnO 4 ~ 

(e) VO ++ + MnOr -> VOr + Mn++ 

+ Zn -> U+ ++ + + Zn++ 



37. The following equations involve oxidation and reduction. They repre- 
sent reactions taking place in the presence of acid. Convert them to balanced 
ionic equations, introducing II + and H 2 O wherever necessary. 

(a) Cr 2 O 7 - + I- -> O+++ + I 2 
(fe) MnO 4 - + H 2 C 2 O 4 -> CO 2 
(c) Cr+++ + Bi0 2 -> Cr 2 O 7 - 



(d) U++++ + MnOr - UO 2 + + + Mn++ 

(e) U0 6 - + H+-lKV-+ + Ot 

(/) I- + Fe(CN) 6 - -> I 2 + Fe(CN) 6 - 

(flf) S 2 0r + I, - S 4 6 - + I- 

38. Write balanced ionic; equations for each of the following reactions 
taking place in acid solution unless otherwise specified. Introduce hydrogen 
ions and water or hydroxyl ions and water wherever necessary, (a) Bichromate 
reduced by sulfite giving chromic salt and sulfate; (6) chromic salt oxidized 
by free chlorine to give chromate and chloride; (c) chromite oxidized in alkaline 
solution with sodium peroxide to chromate; (d) lead peroxide oxidized by 
permanganate giving mangaiious salt and free oxygen; (e) cupric salt and 
metallic aluminum giving aluminum salt and metallic copper; (/) manganous 
salt and chlorate giving a precipitate of manganese dioxide and chlorine 
dioxide gas; (</) cobaltous chloride in alkaline solution with hydrogen peroxide 
to give a precipitate of cobaltic hydroxide. 

39. Express the following reactions in ionic form, and in each case state 
what fraction of the nitric acid employed serves for oxidation: 

(a) Cu + HNO 3 - Cu(NO 3 ) 2 + NO 2 + H 2 O 

(6) Zn + IINO 3 -* Zn(NO 3 ) 2 + NO + H 2 O 

(c) Cu 2 S + HN0 3 -> Cu(NO 3 ) 2 + NO 2 + II 2 SO 4 + H 2 O 

(d) Cu 2 S + HNO 3 -* Cu(NO 3 ) 2 + NO 2 + S + H 2 O 

(e) FeS + HNO 3 -* Fc(N0 3 ) 3 + S + NO + H 2 O 

(/) FeS + HNO 3 -> Pe(NO,) + NO, + H 2 SO 4 + H 2 O 
(g) Sn + HNO 3 + H 2 O ->H 2 SnO 3 + NO + H 2 O 

40. Write balanced ionic equations for the following reactions taking place 
in solution: 

(a) Nitrate + metallic aluminum + sodium hydroxide > aluminate + hydro- 

gen gas + ammonia. 
(6) Bichromate + hydrogen peroxide + acid > chromic salt + oxygen gas. 

(c) Chromate + iodide + acid > chromic salt + iodine. 

(d) Permanganate + sulfite + acid > manganous salt + sulfate. 

(e) Cupric sulfate + iodide > cuprous iodide precipitate + free iodine. 
(/) Cobaltic oxide + acid > cobaltous salt + oxygen gas. 



CHEMICAL EQUATIONS 27 

(g) Manganous salt + permanganate (in neutral solution) > manganese 

dioxide precipitate. 
(h) Mercuric chloride + stannous chloride + hydrochloric acid mercurous 

chloride precipitate + H 2 SnCle 

41. Write balanced ionic equations for the following reactions in which 
nitric acid is used as an oxidizing agent. Introduce H + and H 2 O where 
necessary. 

(a) Cu + HN0 3 - Cu++ + NO 

(6) Zn + (very dilute) HNO 3 - Zn++ + NH^ 

(c) Fe++ + HNO 3 - Fe +++ + NO 

(d) Cu 2 S + HNO 3 -> Cu++ + SO 4 - + NO 2 

(e) Cu 2 S + HNO 3 - Cu ++ + S + NO 

(/) FeSj + HNO 3 -> Fe +++ + SOr + N0 2 
(g) Sn + HNO 3 -. H 2 SnO 3 + NO 
(fc) FesP + HNO 3 - Fe+ ++ + HiPOr + NO 
(0 FeiSi + HNO 3 - Fe +++ + HiSiO 3 + NO 



42. Write balanced ionic equations for the following reactions taking place 
in solution. Introduce H 2 O and other simple constituents wherever necessary. 

(a) Cu(NH,)S0 4 + H 2 S0 4 - CuSO 4 + (NH 4 ) 2 SO 4 

(6) NuAg(CN), + HNO 3 -+ AgCN (precipitate) 

(f) Ag(NH 3 ) 2 Cl + H 2 S -> Ag 2 S (precipitate) 

(d) Cu(NH 3 ) 4 S0 4 + KCN + NII 4 OH -> K 2 Cu(CN) 3 + KCNO 

(e} Cd(NH 3 ) 4 SO 4 + KCN - K 2 Cd(CN) 4 

(/) Co(NH 3 ) 6 Cl 3 + HC1 - CoCl 2 

43. Balance the following molecular equations: (a) Se 2 Cl 2 + H 2 O > 
H 2 SeO 3 + IIC1 + Se; (b) RuQ 4 + 1IC1 - H 3 RuCl 6 + C1 2 + H 2 O; (c) Ag 3 AsO 
+ Zn + II,SO 4 - AsH 3 t + Ag + ZnSO 4 + II 2 O; (d) Ce(IO 3 ) 4 + H 2 (^ 2 O 4 - 
Ce 2 ((J 2 O 4 ) 3 + I 2 + CO 2 ; (e) Fe(CrO 2 ) 2 + Na 2 CQ 3 + O 2 - Fe 2 O 3 + Na 2 CrO 4 H- 
CO (fusion). 



CHAPTER III 

CALCULATIONS BASED ON FORMULAS 
AND EQUATIONS 

15. Mathematical Significance of a Chemical Formula. The 

law of definite proportions states that in any pure compound the 
proportions by weight of the constituent elements are always 
the same. A chemical formula therefore is not only a shorthand 
method of naming a compound and of indicating the constituent 
elements of the compound, but it also shows the relative masses 
of the elements present. 

Thus the formula Na 2 SO 4 (molecular weight = 142.06) indicates 
that for every 142.06 grams of pure anhydrous sodium sulfate 
there are 2 X 23.00 = 46.00 grams of sodium, 32.06 grams of sul- 
fur, and 4 X 16.00 = 64.00 grams of oxygen. The percentage of 

2 X 23 00 
sodium in pure anhydrous sodium sulfate is therefore ^ ' X 

100 = 32.38 per cent. 

16. Formula Weights. A gram-molecular weight of a substance 
is its molecular weight expressed in grams. Thus, a gram-molec- 
ular weight (or gram-mole, or simply mole) of Na 2 SO 4 is 142,06 
grams. A mole of nitrogen gas (N 2 ) is 28.016 grams of the element. 

A formula weight (F.W.) is that weight in grams corresponding 
to the formula of the substance as ordinarily written. In most 
cases it is identical to the gram-molecular weight, but occasionally 
the true molecular weight of a compound is a multiple of the 
weight expressed by the formula as ordinarily written in a chemical 
equation. In practically all the reactions of analytical chemistry, 
however, it can be assumed that the value of the formula weight 
and that of the mole are the same. 

The gram-atom or gram-atomic weight is the atomic weight of 
the element expressed in grams (e.g., 40.08 grams of calcium; 
14.008 grams of nitrogen). A gram-ion is the atomic or formula 
weight of an ion expressed in grams (e.g., 40.08 grams of Ca++; 

62.008 grams of NO 3 ~). 

28 



CALCULATIONS BASED ON FORMULAS AND EQUATIONS 29 

A millimole is one thousandth of a mole; a milligram-atom is 
one thousandth of a gram-atom. 

A formula weight of hydrated ferric sulfate, Fe 2 (SO 4 )3.9H 2 0, 
for example, is 562.0 grams of the salt. It contains 2 gram-atoms 
of iron (= 117.0 grams), 21 gram-atoms of oxygen (= 33G grams), 
9 formula weights (9 F.W.) of water, 3,000 milligram-atoms of 
sulfur, and in solution would give 3 gram-ions of sulfate. 

17. Mathematical Significance of a Chemical Equation. A 
chemical equation not only represents the chemical changes tak- 
ing place in a given reaction but also expresses the relative quan- 
tities of the substances involved. Thus, the molecular equation 

H 2 S0 4 + BaCl 2 -* BaS0 4 + 2HC1 

not only states that sulfuric acid reacts with barium chloride to 
give barium sulfate and hydrochloric acid, but it also expresses 
the fact that every 98.08 parts by weight of sulfuric acid react 
with 208.27 parts of barium chloride to give 233.42 parts of barium 
sulfate and 2 X 36.47 = 72.94 parts of hydrogen chloride, these 
numerical values being the molecular weights of the respective com- 
pounds. These are relative weights and are independent of the units 
chosen. If a weight of any one of the above four substances is 
known, the weight of any or all of the other three can bo calculated 
by simple proportion. This is the basis of analytical computations. 
EXAMPLE I. A sample of pure lead weighing 0.500 gram is 
dissolved in nitric acid according to the equation 

3Pb + 8HN0 3 - 3Pb(N0 3 ) 2 + 2NO + 4H 2 

How many grams of pure HNO 3 are required? How many grams 
of Pb(NO 3 ) 2 could be obtained by evaporating the resulting solu- 
tion to dryness? How many grams of NO gas are formed in the 
above reaction? 
SOLUTION: 

Atomic weight of lead = 207 

Molecular weight of HNO 3 = 63.0 

Molecular weight of Pb(NO 3 ) 2 = 331 

Molecular weight of NO = 30.0 

(3 X 207) grams of Pb react with (8 X 63.0) grams of HNO 3 

(3 X 207) grams of Pb would form (3 X 331) grams of Pb(NO 3 ) 2 , 
and (2 X 30.0) grams of NO 



30 CALCULATIONS OF ANALYTICAL CHEMISTRY 

Hence 0.500 gram of Pb would require 

& v (\1 O 
0.500 X tC = - 405 g fam of HNO ' 

o X 

and would form 

and 

9 v n n 
0.500 X *m = 0.0483 gram of NO 

O X 



0.500 X - = - 799 g ram of Pb(N0 3 ) 2 

o X 



EXAMPLE II. How many grams of H 2 S would be required to 
precipitate the lead as lead sulfide from the above solution? How 
many milliliters of H2S under standard temperature and pressure 
would be required for the precipitation? (A gram-molecular 
weight of a gas under standard conditions occupies 22.4 liters. 
See Sec. 110.) 
SOLUTION: 

Pb++ + H 2 S - PbS + 2H+ 

Atomic weight of lead = 207 

Molecular weight of H 2 S = 34.1 

207 grams of Pb 4 " 4 " require 34.1 grams of H 2 S 

Hence 0.500 gram of Pb + + requires 

0.500 X ~ *? 0.0822 gram of H 2 S. Ans. 

34.1 grams of H 2 S occupy 22,400 ml. under standard conditions 

OR99 
Volume of H 2 S - - X 22,400 = 54.1 ml. 



EXAMPLE III. In the reaction expressed by the equation: 



2Ag 2 C0 3 -> 4Ag + 2 + 2C0 2 

(a) how many gram-atoms of silver can be obtained from 1 F.W. 
of silver carbonate, (6) how many gram-atoms of silver can be 
obtained from 1.00 gram of silver carbonate, (c) how many grams 
of silver carbonate are required to give 3.00 grams of oxygen 
gas, (d) how many moles of gas (CO 2 + 2 ) are produced from 
50.0 grams of silver carbonate, and (e) how many milliliters 
of gas (C0 2 + O 2 ) are produced from 1 millimole of silver car- 
bonate? 



CALCULATIONS BASED ON FORMULAS AND EQUATIONS 31 

SOLUTION: 

(a) 2 F.W. Ag 2 CO 3 > 4 gram-atoms Ag 



1 F.W. Ag 2 CO 3 2 gram-atoms Ag. 
(6) Molecular weight Ag 2 CO 3 = 276 

1 fin 1 HA 

1.00 gram Ag 2 CO 3 = -^~- g = ^ = 0.00363 F.W. Ag2C0 8 

0.00363 X 2 = 0.00726 gram-atom Ag. Ans. 

(c) 2 moles Ag 2 C0 3 (= 2 X 276 = 552 grams) give 1 mole 2 

(= 32 grams) 

KKO 

3.00 grams O 2 = 3.00 X -^- = 51 .7 grams Ag 2 C0 3 . Ans. 

d 

(d) 2 moles Ag 2 CO 3 (= 552 grams) > 3 moles (O 2 + CO 2 ) 

50 
50.0 grams Ag 2 C0 3 = -^ X 3 = 0.272 mole gas. Ans. 



(e) 1 mole Ag 2 CO 3 > 1^ moles gas 

1 mole gas (standard temperature and pressure) = 22,400 mL 

1 mfflimole Ag 2 CO 3 = 1^ X 

= 33.8 ml. of gas. Ans. 
EXAMPLE TV. In the reaction expressed by the equation 

MnO 2 + 2NaCl + 3H 2 SO 4 -> MriSO 4 + 2NaHSO 4 + C\ 2 + 2H 2 O 
or 



MnO 2 + 2C1- + 6H+ ~> Mn++ + C1 2 + 2H 2 O 

(a) how many gram-ions of Mn++ can be obtained from 1 milli- 
mole of MnO 2 , (6) how many grams of MnSO 4 can be obtained 
from 5.00 grams of MnO 2 , (c) how many millimoles of MnO 2 are 
required to give 100 ml. C1 2 (standard conditions), and (d) if 
1.00 gram of MnO 2 , 1.00 gram of NaCl, and 5.00 grams of H 2 SO 4 
are used, which is the limiting reagent, and how many milliliters 
of C1 2 (standard conditions) are evolved? 
SOLUTION : 

(a) 1 mole MnO 2 > 1 gram-ion Mn++ 

1 millimole MnO 2 > 0.001 gram-ion Mn 4 " 1 ". Ans. 

(b) 1 mole Mn0 2 (= 86.9 grams) -> 1 mole MnS0 4 (= 151 grams) 

151 
5.00 grams MnO 2 = 5.00 X SJTS 

oO.u 

= 8.69 grams MnSO 4 . Ans. 



32 CALCULATIONS OF ANALYTICAL CHEMISTRY 



(c) 100 ml. C1 2 = = 4.47 millimoles C1 2 

ZZ.4 

1 millimole Cl2 = 1 millimole MnO 2 
100 ml. C1 2 = 4.47 millimoles MnO 2 . Ans. 

i no i (\r\ 

(d) 1.00 gram MnO 2 = ~~ = = 0.0115 mole 



00 1 00 

1.00 gram NaCl - ^-, = ~ = 0.0171 mole 

oo.o 



5.00 grams H 2 SO 4 = ^^ = sir? = 0.0510 mole 

X12OU4 yo. 1 

According to the equation these substances react in the molar 
ratio of 1:2:3, or 0.0115:0.0230:0.0345. The NaCl is therefore 
the limiting reagent and the other two are in excess. 

2 moles NaCl -> 1 mole C1 2 = 22,400 ml. C1 2 
0.0171 mole NaCl -> ' * 71 X 22,400 

Zi 
= 192 ml. C1 2 . Ans. 

Problems 

44. How many grams of potassium and of carbon are contained in (a) 0.211 
gram of K4Fe(CN) 6 .3II 2 O; (6) 1 F.W. of KHC 4 H 4 O 6 ? 

Ans. (a) 0.0782 gram, 0.0360 gram; (b) 39.1 grams, 48.0 grams. 

46. A certain weight of lead phosphate, Pb 3 (PO 4 ) 2 , contains 0.100 gram of 
lead. How many grams of phosphorus are present? What is the weight of the 
lead phosphate? What is the percentage of oxygen present? 

Ans. 0.00997 gram. 0.131 gram. 15.8 per cent. 

46. How many grams of oxygen are present in 1.00 gram of each of the 
following: (a) Fe 2 O 3 , (b) BaSO 4 , (c) Fe(NO 3 ) 3 .6H 2 O? 

Ans. (a) 0.300 gram, (b) 0.275 gram, (c) 0.686 gram. 

47. What is the percentage by weight of sulfur in each of the following: 
(a) BiA, (b) Na 2 S 2 O3.5II 2 O, (c) K 2 SO 4 .A1 2 (SO 4 ) 3 .24H 2 O? 

Ans. (a) 18.7 per cent, (b) 25.8 per cent, (c) 13.5 per cent. 

48. Ignition of anhydrous magnesium ammonium phosphate forms mag- 
nesium pyrophosphate according to the equation: 2MgNH 4 PO 4 > Mg 2 P 2 O 7 + 
2NH 3 + H 2 O. Calculate: (a) number of formula weights of Mg 2 P 2 O 7 pro- 
duced from 1.00 F.W. of MgNH 4 PO 4 , (b) number of grams of NH 3 produced 
at the same time, (c) number of milliliters of NH 3 (standard conditions) 
accompanying the formation of 1 millimole of Mg 2 P 2 O 7 . 

Ans. (a) 0.500, (b) 17.0 grams, (c) 44.8 ml. 



CALCULATIONS BASED ON FORMULAS AND EQUATIONS 33 

49. What is the weight of the constituent elements in 0.717 gram of AgN0 3 ? 
Ans. Ag = 0.455 gram, N = 0.059 gram, O = 0.203 gram. 

60. Calculate the number of pounds of materials theoretically necessary for 
the preparation of 1.00 pound of (a) KOH from CaO and K 2 CO 3 , (6) BaSO 4 
from Na2S04.10II 2 and BaCl 2 .2H 2 O. 

Ans. (a) CaO = 0.500 pound, K 2 CO 3 = 1.23 pounds. 

(&) Na 2 SO 4 .10H 2 O = 1.38 pounds, BaCl 2 .2H 2 O = 1.04 pounds. 

51. Balance the following equation and also write it as a balanced ionic 
equation: A1 2 (SO 4 )3 + BaCl 2 - A1C1 3 + BaSO 4 . Calculate from it the follow- 
ing: (a) number of gram-ions of Al" 1 " 1 "*" contained in 1 gram-mole of A1 2 (SO 4 )3, 
(b) number of gram-ions of Ba ++ reacting with 1 .00 gram of A1+ *~+, (c) number 
of grains of BaSO 4 obtainable from 2.00 grams of A1 2 (SO 4 ) 3 .18II 2 O, (d) number 
of grams of BaSO 4 produced by mixing solutions containing 3.00 grams of 
A1 2 (SO 4 ) S and 4.00 grains of BaCU. 

Ans. (a) 2, (b) 0.0556, (c) 2.10 grams, (d) 4.48 grams. 

52. From the reaction: 4FeSa + 1 10 2 -* 2Fe 2 O 3 + 8SO 2 , calculate the fol- 
lowing: (a) number of moles of FeS 2 required to form 1 F.W. of Fe 2 O 3 , (b) num- 
ber of grams of oxygen required to react with 2.00 moles of FeS 2 , (c) number 
of millimoles of SO 2 equivalent to 0.320 gram of O 2 , (d) volume of SO 2 (standard 
conditions) accompanying the formation of 0.160 grain of Fe 2 O 3 . 

Ans. (a) 2, (b) 176 grams, (c) 7.27, (d) 89.6 ml. 

53. Complete and balance the following ionic equation for a reaction taking 
place in the presence of acid: Fe+ + + MnO 4 ~" ^Fe ++ ^ + Mn 4 +. Calculate 
from it the following: (a) number of gram-ions of Mn ++ produced from 1 gram- 
ion of Fe++, (6) number of millimoles of Fe 2 (SO 4 ) 3 .9H 2 O obtainable if 1 milli- 
mole of KMnO 4 is reduced, (c) decrease in the number of gram-ions of H 4 " 
accompanying the formation of 1.00 gram of Fe 4 " 1 "*, (d) number of grams of 
Fe 2 (SO 4 ) 3 obtainable by mixing solutions containing 1.00 gram of FeSO 4 .7II 2 O, 
0.100 gram of KMnO 4 , and 1.00 gram of H 2 SO 4 . 

Ans. (a) y 5 , (b) 2%, (c) 0.0286, (d) 0.633 gram. 

64. What weight of NH 3 is required to dissolve 0.120 gram of AgCl accord- 
ing to the equation: AgCl + 2NH 3 -> Ag(NH 3 ) 2 + + Cl~? 
Ans. 0.0285 gram. 

66. How many grams of H 2 S are required to precipitate the bismuth as 
Bi 2 S 3 from an acid solution containing 100 mg. of dissolved bismuth? 

Ans. 0.0244 gram. 

66. How many grams of H 2 SO 4 are required to dissolve 0.636 gram of 
metallic copper according to the equation: Cu + 2H 2 SO 4 - CuSO 4 + SO 2 + 
2H 2 O? How many milliliters of gas are evolved (measured under standard 
conditions)? 

Ans. 1.96 grams. 224 ml. 



34 CALCULATIONS OF ANALYTICAL CHEMISTRY 

67. How many grains of anhydrous chromic chloride (CrCl 3 ) could be ob- 
tained from 100 mg. of K 2 Cr 2 O7 after reduction by H 2 S in the presence of HC1: 
Cr 2 O 7 ~ + 3H 2 S + 8H+ -> 2Cr+++ + 38 + 7II 2 O? How many grams and how 
many milliliters (standard conditions) of H 2 S would be required? 

Ans. 0.0723 gram, 0.0347 gram, 22.8 ml. 



68. How many grams of chromium are present in 0.250 gram of K 2 Cr 2 C>7? 
What is the percentage of potassium in this compound? 

69. What weight of alum, K 2 SO 4 .A1 2 SO4.24II 2 O, contains 0.200 gram of 
aluminum? What is the percentage of oxygen in the compound? 

60. What weight of sulfur is present in an amount of Na 2 S 2 O 3 that contains 

(a) 318 mg. of sodium, (fe) 1.00 gram-atom of oxygen? 

61. How many grams of nitrogen are present in 1.00 gram of each of the 
following: (a) NH 3 , (b) Pb(NO 3 ) 2 , (c) FeSO 4 .(NH4) 2 SO 4 .6H 2 O? 

62. What is the percentage of oxygen in each of the following: (a) H 2 O, 

(b) FeSO 4 .7H 2 O, (c) K 2 SO 4 .Cr 2 (SO 4 ) 3 .24H 2 O? 

63. Ignition of bismuth basic carbonate takes place according to the follow- 
ing equation: 2BiOIICO 3 -> BiA, + 2CO 2 + H 2 O. Calculate the following: 

(a) number of formula weights of Bi 2 O 3 produced from 1 F.W. of the carbonate, 

(b) number of millimoles of OO 2 accompanying the formation of 1.00 gram of 
Bi 2 O?, (c) volume of CO 2 (standard conditions) formed from 0.0200 gram of 
BiOIICOs, (d) volume of gas (CO 2 + water vapor) accompanying the forma- 
tion of 1.00 millimole of Bi 2 O 3 . 

1 64. Convert the following to balanced molecular and ionic equations: 
eC! 3 + AgNO 3 ->Fe(NO 3 ) 2 + AgCl. Calculate from them the following: 
(a) number of formula weights of AgCl obtainable from 1 F.W. of FeCl 3 , 
'(b) number of gram-ions of Fe *' ++ produced per millimole of AgCl, (c) number 
-of grams of Fe(NO 3 )3.6H 2 O obtainable if 1.00 gram-molecular weight of 
AgNO 3 is used up, (d) number of grams of AgCl obtained by mixing solutions 
containing 0.700 gram of FeCl 3 and 0.600 gram of AgNO 3 . How many grams 
of which reactant are left over? 

66. Assuming that the reaction for the fusion of the mineral chromite with 
Na 2 CO 3 + NaNO 3 takes place according to the equation: 10Fe(CrO 2 ) 2 + 
13Na 2 CO 3 + 14NaNO 3 - 5Fe 2 O ? + 20Na 2 CrO 4 + 7N 2 + 13CO 2 , how many 
millimoles and how many milliliters (measured at 760 mm. and 0C.) of gas 
are formed from that weight of Fe(CrO 2 ) 2 containing (a) 1.00 gram-atom of 
Cr, (6) 1.00 gram of Cr? 

66. How many milligrams of NH 3 are required to react with 27.2 mg. of 
Hg 2 Cl 2 according to the equation: Hg 2 Cl 2 + 2N T H 3 ->IIgNH 2 Cl + Hg + 
NH 4 + + Cl~? How many grams of free mercury would be formed? 

67. How many grams and how many milliliters (standard conditions) of 
H 2 S are required to precipitate the arsenic from an acid solution containing 
0.100 gram of Na 3 AsO 3 : 2AsO 3 a + 3H 2 S + 6H + -> As 2 S 3 + 6H 2 O? How 
many grams of the sulfide would be formed? 



CALCULATIONS BASED ON FORMULAS AND EQUATIONS 35 

68. Balance the following equation: MnOr + Fe ++ + H+ -> Mn++ + 
Fe + ++ + H 2 O and calculate from it the number of grains of FeS04.7H 2 O re- 
quired to reduce that weight of KMnO4 that contains 0.250 gram of Mn. 

69. Balance the following equation: Cr 2 7 ~ + Fe++ + H+ -> Cr+++ + 
Fe +f+ + H 2 O. If 1.00 gram-molecular weight of K 2 CrO 4 is dissolved in water 
and the solution acidified (2Cr04 = + 2H+ > Cr 2 07 = " + H 2 0), how many grams 
of FeSO 4 .(NH 4 ) 2 SO4.6H 2 O would be required to reduce the chromium in the 
resulting solution? 

70. When used for the oxidizing effect of its nitrate, which is the more 
economical reagent, potassium nitrate at 65 cents per pound or sodium nitrate 
at 50 cents per pound? How much is saved per pound of the more economical 
reagent? 



CHAPTER IV 
CONCENTRATION OF SOLUTIONS 

18. Methods of Expressing Concentration. Solution reagents 
used in analytical chemistry are usually either (1) laboratory 
reagents the concentrations of which need be known only ap- 
proximately, or (2) titration reagents the concentrations of which 
must be known to a high degree of precision. There are several 
ways of expressing concentration and it is important to have a 
clear understanding of just what is meant in each case. In ana- 
lytical work the following methods of expressing concentration are 
most commonly used. 

19. Grams per Unit Volume. By this method a concentra- 
tion is expressed in terms of the number of grams (or milligrams) 
of solute in each liter (or milliliter) of solution. A 5-gram-per- 
liter solution of sodium chloride is prepared by dissolving 5 grams 
of the salt in water and diluting to one liter (not by adding one 
liter of water to the salt). 

This method is simple and direct but it is not a convenient 
method from a stoichiometric point of view, since solutions of the 
same concentration bear no simple relation to each other so far 
as volumes involved in chemical reactions are concerned. Chemi- 
cal substances enter into reaction upon a mole-to-mole basis and 
not upon a gram-to-gram basis. 

20. Percentage Composition. This method is on a percentage 
by-weight basis and expresses concentration in terms of grams of 
solute per 100 grams of solution. A 5 per cent of sodium chloride 
is made by dissolving 5 grams of the salt in 95 grams of water, 
which of course gives 100 grams of solution. 

21. Specific Gravity. The specific gravity of the solution of a 
single solute is a measure of the concentration of the solute in the 
solution. Although occasionally used in analytical chemistry, it 
is a cumbersome method, since it necessitates consulting a table 
in order to determine the percentage-by-weight composition. 
Tables of specific gravities of common reagents are found in the 

36 



CONCENTRATION OF SOLUTIONS 37 

handbooks and other reference books of chemistry. Tables cov- 
ering the common acids and bases are also in the Appendix of this 
text. Here it will be found, for example, that hydrochloric acid 
of specific gravity 1.12 contains 23.8 grams of hydrogen chloride 
in 100 grams of solution. 

22. Volume Ratios. Occasionally in analytical work the con- 
centration of a mineral acid or of ammonium hydroxide is given 
in terms of the volume ratio of the common concentrated reagent 
and water. Thus HC1 (1:3) signifies a solution of hydrochloric 
acid made by mixing one volume of common, concentrated 
hydrochloric (sp. gr. about 1.20) with three volumes of water. 
vSimilarly H 2 SO 4 (l"-3) signifies a solution made by mixing one 
volume of the commonly used concentrated sulfuric acid (sp. 
gr. 1.84) with three parts by volume of water. This method of 
expressing concentrations is cumbersome, particularly in work 
where subsequent calculations involving the solutions are to be 
made. 

23. Molar and Formal Solutions. A molar solution is one con- 
taining a gram-mole of substance dissolved in a liter of solution. 
This is usually identical to a formal solution which contains a 
formula weight of substance in a liter of solution (see Sec. 16). 
A gram-molecular weight of substance dissolved in a liter of water 
does not constitute a molar solution, for the resulting solution 
does not occupy a volume of exactly a liter. 1 A liter of molar 
(M) sulfuric acid solution contains 98.08 grams of H 2 SO 4 ; a liter 
of half-molar (J^M, 0.5M, or M/2) sulfuric acid solution contains 
49.04 grams of H 2 S04. In this particular case 98.08 grams of H 2 S04 
does not mean 98.08 grams of the ordinary concentrated sulfuric 
acid, but of hydrogen sulfate. The concentrated acid contains 
about 96 per cent of the latter. 

Since 1 mole of hydrochloric acid reacts with 1 mole of sodium 
hydroxide, a certain volume of sodium hydroxide solution will be 
exactly neutralized by an equal volume of hydrochloric acid of the 
same molar concentration, or twice the volume of hydrochloric 
acid of one-half the molar concentration of the sodium hydroxide. 

1 Solutions containing a gram-molecular weight of substance dissolved in 
1,000 grams of water are useful in computations involving certain physico- 
chemical phenomena. Such solutions are often referred to as molal solutions, 
but this standard is not used in general analytical work. 



38 CALCULATIONS OF ANALYTICAL CHEMISTRY 

One molecule of hydrogen sulfate will neutralize 2 molecules of 
sodium hydroxide. 

H 2 S0 4 + 2NaOH -> Na 2 S0 4 + 2H 2 

To neutralize a certain volume of sodium hydroxide solution, only 
one-half that volume of sulfuric acid of the same molar concentra- 
tion would be required. Volumetric calculations are therefore 
greatly simplified when concentrations are expressed in terms of 
moles of substance per unit volume of solution; for, when so ex- 
pressed, the volumes of reacting solutions of the same molar con- 
centration, although not necessarily equal, bear simple numerical 
relationships to each other. 

EXAMPLE. What volume of 0.6380 M potassium hydroxide 
solution will neutralize 430.0 ml. of 0.4000 M sulfuric acid? 
SOLUTION: 

1 mole H 2 S0 4 ~ 2 moles KOH 

430.0 ml. of 0.4000 molar solution contains 

x 0.4000 = 0.1720 mole H 2 SO 4 

0.1720 mole H 2 SO 4 o 0.3440 mole KOH 
1 ml. KOH contains 0.0006380 mole KOH 

Volume required = ^ r^r^ooT^ = 539.3 ml. Ans. 
U.UUUoooU 

24. Equivalent Weight and Normal Solution. The equivalent 
weight of an element or compound is that weight equivalent in 
reactive power to one atomic weight of hydrogen. The milli- 
equivalent weight is one thousandth of the equivalent weight. The 
gram-equivalent weight is the equivalent weight expressed in grams; 
the gram-milliequivalent weight is the milliequivalent weight ex- 
pressed in grams. 1 The application of gram-equivalent weights 
to various types of chemical reactions will be taken up in detail 
in Part III, but simple cases, applying particularly to qualitative 
analysis, will be considered briefly here. 

1 The equivalent weight of a substance, like the atomic or molecular weight, 
is merely a number without a unit of weight; the gram-equivalent weight is a 
definite number of grams. However, when the connotation is clear, the terms 
" equivalent weight" and "milliequivalent weight" are frequently used to 
signify gram-equivalent weight and gram-milliequivalent weight, respectively. 



CONCENTRATION OF SOLUTIONS 39 

The gram-equivalent weight of an acid, base, or salt involved 
in a simple metathesis such as a neutralization or precipitation 
is that weight in grams of the substance equivalent in neutralizing 
or precipitating power to 1 gram-ion of hydrogen (i.e., 1.008 grams 
of H+). 

A normal solution contains 1 gram-equivalent weight of solute 
in 1 liter of solution, or 1 gram-milliequivalent weight in 1 milli- 
liter of solution. The normality of a solution is its relation to a 
normal solution. A half-normal solution therefore contains in a 
unit volume one-half the weight of solute contained in its normal 
solution, and this weight may be expressed as 0.5 N, ^ N, or 
N/2. The concentration of a normal solution is expressed simply 
as N. 

Since the concentrations of solutions used in precise volumetric 
analysis are usually found experimentally, the concentrations can- 
not often be expressed by whole numbers or by simple fractions. 
They are more likely to be expressed as decimal fractions, e.g., 
0.1372 N. 

26. Simple Calculations Involving Equivalents, Milliequivalents, 
and Normality. The use of equivalents, milliequivalents, and 
normality is so extensive in analytical chemistry and the terms 
are so fundamental that a clear understanding of them is essential 
at this time. More detailed discussions applying particularly to 
quantitative analysis will be given in Part III. 

Let us consider here only the simplest reactions between com- 
mon acids, bases, and salts, and as an example let us take sulfuric 
acid. The molecular weight of II 2 SO 4 is 98.08. A mole, or gram- 
molecular weight, of H 2 S0 4 is 98.08 grams, and a molar solution 
of the acid therefore contains this amount of pure hydrogen sul- 
fate in a liter of solution. Since 98.08 grams of H 2 SO4 has a neu- 
tralizing power equivalent to 2 gram-atoms (2.016 grams) of hy- 
drogen as an ion, the gram-equivalent of H 2 SO4 as an acid is 
98.08/2 = 49.04 grams, which is equivalent in neutralizing power 
to 1 gram-atom (1.008 grams) of hydrogen as an ion. The gram- 
milliequivalent weight is 0.04904 gram. A normal solution of sul- 
furic acid therefore contains 49.04 grams of ILSO4 in a liter of 
solution, or 0.04904 gram of H 2 S04 in a millimeter of solution. 
A 1 molar solution of sulfuric acid is 2 normal; a 1 normal solu- 
tion of sulfuric acid is ]/2 molar. 



40 



CALCULATIONS OF ANALYTICAL CHEMISTRY 



Sodium hydroxide is a base with a molecular weight of 40.00. The 
gram-equivalent weight of NaOH is 40.00 grams, since this amount 
is neutralized by 1.007 grams of 11+ . A normal solution of NaOH 
contains 40.00 grams in a liter of solution and is likewise 1 molar. 



ACIDS 



- =36.47 g 



H 2 S0 4 


= 49.04 g. 



^2^=60.05 g. 



These weights are all 
equivalent to 




-y2-=40.03g. 



Na 3 P0 4 


-= 54.66 g. 



FIG. 1. Gram-equivalent weights of some acids, bases, and salts. 

The gram-equivalent weight of a simple salt is determined in 
the same way as that of an acid or base, namely by reference to 
1.008 grams of 11+ as a standard. In the case of the salt of a 
metal, the equivalent weight is ordinarily the molecular weight 
of the salt divided by the total oxidation number represented by 
the atoms of metal in the formula. 

The equivalent weights of a few acids, bases, and salts are 
shown in Fig. 1. Since each of these amounts is equivalent to the 
same standard, they are mutually equivalent to one another. 



CONCENTRATION OF SOLUTIONS 41 

In each case the specified amount when dissolved in one liter 
of solution will produce a 1 normal solution. 

It follows that 1 liter of 1 N HC1 will neutralize 1 liter of 1 N 
NaOH, or 1 liter of 1 N Ba(OH) 2 , or 1 liter of any one-normal 
base. One liter of 1 N H2SO4 will also neutralize 1 liter of any 
one-normal base. More generally, a certain volume of any acid 
will neutralize the same volume of any base of the same normality. 

Similarly, 1 liter of 1 N AgNO 3 will precipitate the chloride 
from 1 liter of 1 N NaCl or 1 liter of 1 N BaCl 2 , and the latter 
will just precipitate the sulfate from 1 liter of 1 N Na 2 SO4 or 1 liter 
of i N Fe 2 (SO 4 ) 3 . 

We found that when two solutions of equal molarity react, the 
volumes are in simple ratio to each other. But when two solu- 
tions of equal normality react, the volumes of the solution are 
equal. 

Since volumes of reagents in analytical chemistry are usually 
measured in milliliters rather than in liters, it is more convenient 
to consider a normal solution as containing I gram-millicquivalent 
weight per milliliter. Hence the number of gram-milliequivalent 
weights present in a solution can be found from the simple rela- 
tionship : 

Number of milliliters X normality = 

number of gram-milliequivalent weights 
or 

ml. X N = number of me. vvts. 

(See footnote, p. 38) 

Thus, 2.00 ml. of 0.00 N HOI contain 12.0 millicquivalent weights, 

I1C1 
or 12.0 X T-TWWX = 0.438 gram of hydrogen chloride. This will 

L,UUU 

exactly neutralize 12.0 milliequivalents of any base, for example, 
4.00 ml. of 3.00 N NaOH, or 4.00 ml. of 3.00 N Na 2 CX) 3 , or 80.0 ml. 
of0.150NBa(OH) 2 , etc. 

It follows that when solutions A and B mutually interact to a 
complete reaction, 

ml.,i X N,i = m\. B X N* 

EXAMPLE I. What is the approximate molarity and normality 
of a 13.0 per cent solution of H 2 SO 4 ? To what volume should 
100 ml. of the acid be diluted in order to prepare a 1.50 N solution? 



42 CALCULATIONS OF ANALYTICAL CHEMISTRY 

SOLUTION: From specific gravity table in the Appendix, the 
specific gravity of the acid is 1.090. 
1 liter weighs 1,090 grams 

1 liter contains 1,090 X 0.130 = 142 grams H 2 S0 4 
1 mole II 2 SO 4 = 98.08 grams 

Molarity of solution = 142/98.08 = 1.45 M. Ans. 
1 gram-equivalent H 2 SO 4 = H 2 SO 4 /2 = 49.04 grams 
Normality of solution = 142/49.04 = 2.90 N. Ans. 
100 ml. contain 290 milliequivalents H 2 SO 4 
After dilution x ml. of 1.50 N contain 290 milliequivalents 

x X 1.50 = 290 

x = 193 ml. Ans. 

EXAMPLE II. A solution contains 3.30 grams of Na 2 CO 3 .10H 2 O 
in each 15.0 ml. What is its normality? What is its molarity? 
With how many milliliters of 3.10 N acetic acid, HC 2 H 3 O 2 , will 
25.0 ml. of the carbonate react according to the equation: 2H+ + 
C0 8 - - H 2 O + C0 2 ? With how many milliliters of 3.10 N H 2 S0 4 
will 25.0 ml. of the carbonate react? 
SOLUTION : 

Molecular wt. Na 2 CO 3 .10H 2 O = 286 



Equivalent wt. Na2CO 8 .10H 2 O = = 143 

Zt 

Milliequivalent weight = 0.143 

3 30 

Solution contains y^ = 0.220 gram per ml. 
JLo.U 

1 normal solution would contain 0.143 gram per ml. 
Normality = ' . Q = 1.54 N. Ans. 



Molarity = = 0.77 M. Ans. 



x X 3.10 = 25.0 X 1.54 

x = 12.4 ml. HC 2 H 3 O 2 . Ans. 
= 12.4 ml. H 2 SO 4 . Ans. 

EXAMPLE III. (a) A 0.100 M solution of aluminum sulfate, 
A1 2 (S0 4 ) 3 , would be of what normality as an aluminum salt? 
(6) What normality as a sulfate? (c) How many milliequivalents 
of the salt are contained in each milliliter? (d) What volume of 



CONCENTRATION OF SOLUTIONS 43 

6.00 N NH 4 OH would be required to react with the aluminum 
in 35.0 ml. of the salt solution according to the equation: A1+++ + 
3NH 4 OH -> A1(OH) 8 + 3NH 4 + ? (e) What volume of 6.00 N 
BaCl2.2H 2 O solution would be required to precipitate the sulfate 
from 35.0 ml. of the solution? (/) How many grams of BaCl 2 .2H 2 O 
are contained in each milliliter of the above solution? 
SOLUTION : 

(a) 1 mole A1 2 (SO 4 ) 3 = 6 equivalents (2 A1+++ o 6H+) 

0.100 molar = 0.600 normal as Al salt. Ans. 
(6) = 0.600 normal as sulfate. Ans. 

(c) 0.600 milliequivalent per milliliter. Ans. 

(d) ml.A X N^ = ml.* X N* 

x X 6.00 = 35.0 X 0.600 

x = 3.50 ml. Ans. 

(e) ml.A X N^ = ml* X N* 

x X 6.00 = 35.0 X 0.600 

x = 3.50 ml. Ans. 



(/) 0.00 X = 0.732 gram. Ans. 

Problems 

71. What fraction of the molecular weight represents the milliequivalent 
weight in the case of each of the following acids, bases, and salts: (a) H 2 SiF 6 , 
(fc) H 3 As0 4 , (c) H 4 P 2 7 , (d) Th0 2 , (e) (NII 4 ) 2 SO 4 , (/) Zn 3 (AsO 3 ) 2 ? 

Ans. (a) 1/2,000, (b) 1/3,000, (c) 1/4,000, (d) 1/4,000, (e) 1/2,000, 
(/) 1/6,000. 

72. How many grams of K 2 S0 4 are contained in 50.0 ml. of 0.200 N solu- 
tion? How many millimoles of K 2 SO 4 are present? 

Ans. 0.872 gram. 5.00 millimoles. 

73. A solution of H 2 SO 4 has a specific gravity of 1.150. What is the 
normality of the solution? 

Ans. 4.90 N. 

74. What is the normality of a solution of NH 4 OH having a specific gravity 
of 0.900? How many milliliters of 13.0 N H 2 SO 4 would be neutralized by 
15.0 ml. of the NH 4 OH? To what volume should 250 ml. of the 13.0 N H 2 SO 4 
be diluted to make a solution that is 5.00 molar? 

Aiis. 15.0 N. 17.3ml. 325ml. 

76. A 30 per cent solution of H 3 PO 4 has a specific gravity of 1.180. What 
is its normality as an acid assuming partial neutralization to form HPO 4 "? 
What is its molar concentration? 

Ans. 7.22 N. 3.61 M. 



44 CALCULATIONS OF ANALYTICAL CHEMISTRY 

76. How many grams of SrCk.GHgO are required to prepare 500 ml. of 
0.550 N solution? What is the molarity of the solution? How many milliliters 
of 1.00 N AgNOs would be required to precipitate the chloride from 20.0 ml. 
of the strontium chloride solution? 

Ans. 36.6 grams. 0.275 M. 11.0 ml. 

77. How much water must be added to 50.0 ml. of a 0.400 N solution of 
Cr 2 (SO 4 )3.18H 2 O in order to make it 0.0500 molar? How many milliliters of 
0.200 N NH 4 OH would be required to precipitate all the chromium as Cr(OH) 3 
from 20.0 ml. of the original undiluted solution? 

Ans. 16.7ml. 40.0ml. 

A piece of aluminum weighing 2.70 grams is treated with 75.0 ml. of 
(sp. gr. 1.18 containing 24.7 per cent H 2 SO 4 by weight). After the metal 
is completely dissolved (2A1 + 6H+ > 2A1+++ + 3H 2 ) the solution is diluted 
to 400 ml. Calculate (a) normality of the resulting solution in free sulfuric 
acid, (b) normality of the solution with respect to the aluminum salt it contains, 
(c) total volume of 6.00 N NH 4 OH required to neutralize the acid and precipi- 
tate all the aluminum as A1(OH)3 from 50.0 ml. of the solution. 
Ans. (a) 0.365 N, (6) 0.750 N, (c) 9.30 ml. 




79. What is the gram-equivalent weight of each of the following acids, bases 
and salts: (a) II 2 C2O 4 .2H 2 O, (6) H 3 PO 3 , (c) CaO, (d) Fe 2 3 , () SnCU, 
(/) (Ca 3 (P0 4 ) 2 ? 

80. How many milliliters of 2.30 M H 2 SO 4 would be neutralized by 15.8 ml. 
of 3.20 M NaOH? How many milliliters of 4.60 N H 2 C 2 O 4 .2H 2 O solution 
would be neutralized by 10.0 ml. of 5.10 N NaOH? By 10.0 ml. of 5.10 N 
Ba(OH) 2 ? How many milliliters of 4.60 N HC 2 H 3 O 2 would be neutralized by 
10.0 ml. of 5. ION NaOH? 

81. To what volume must 25.0 ml.of HC1 (sp. gr. 1.100) be diluted in order 
to make a solution of HC1 with a specific gravity of 1.04? How many milli- 
liters of 0.500 N Ba(OH) 2 would be required to neutralize 40.0 ml. of the re- 
sulting diluted solution? 

82. A 12.0 per cent solution of H 2 C 2 O 4 .2H 2 O has a specific gravity of 1.04. 
What is the normality of the solution as an acid? How many milliliters of 
3.00 molar KOH would be neutralized by 18.0 ml. of the acid? 

83. How many milliliters of 0.500 N BaCl 2 solution would be required to 
precipitate all the sulfate from 10.0 millimoles of FeSO 4 .(NH 4 ) 2 SO 4 .6H 2 O? 
How many milliliters of 0.100 N AgNO 3 would be required to precipitate the 
chloride from 8.30 ml. of the barium chloride solution? 

84. What is the approximate normality of a solution of nitric acid marked 
"HN0 3 1:4"? How many milliliters would theoretically be required to react 
with 0.028 F.W. of Fe 2 O 3 to form Fe(NO 3 ) 3 ? 

86. A solution of H 3 PO 4 contains 0.500 millimole per milliliter. What is the 
normality of the solution as a phosphate? How many milliliters of 1.20 N 



CONCENTRATION OF SOLUTIONS 45 

KOH would be required to form KH 2 PC>4 with 5.00 ml. of the phosphoric acid? 
To what volume must 25.0 ml. of the original acid be diluted in order to make 
the solution 1.10 N as a phosphate? 

86. How many grams of FeCl 3 are contained in 25.0 ml. of 0.520 N ferric 
chloride solution? How many millimoles of FeCl 3 .6H 2 O could be obtained by 
evaporating the solution to dryness? How many milliliters of 0.200 N NH 4 OH 
are required to react with 25.0 ml. of the ferric chloride solution to precipitate 
Fe(OH) 8 ? 



CHAPTER V 
EQUILIBRIUM CONSTANTS 

26. Law of Mass Action. In simple terms the law of mass 
action may be expressed as follows: The rate of reaction between 
two or more interacting substances in a mixture is proportional to 
the product of the prevailing active concentrations of the substances. 

A great many of the reactions of analytical chemistry are re- 
versible reactions. This means that the products of a given reac- 
tion interact, at least to some extent, to give the initial substances. 
Consider a general reversible reaction between substances A and 
B at a given temperature to give substances C and D according 
to the following equation: 

A + B <= C + D 

At the start of the reaction, only substances A and B are present. 
These react at a certain rate to give C and D and, as the latter 
are produced, the concentrations of A and B decrease. According 
to the law of mass action, the rate of the reaction between A and 
B at any given moment is proportional to the prevailing concen- 
trations of A and B at that moment. In symbols this may be 
expressed as follows: 

Rate of reaction between A and B = k'[A][B] 

where [A] and [B] are the prevailing molar concentrations of A 
and B, respectively, and k' is a constant at a given temperature. 
As the concentrations of substances C and D increase, these sub- 
stances in turn react at a constantly increasing rate to produce 
A and B. The rate of this reaction at any moment is proportional 
to the product of the prevailing concentrations of C and D. 

Rate of reaction between C and D = k"[C][D] 

When equilibrium has been established, these two rates are equal. 
Hence, 

[C][D] _ K = TC 

[A][B] ~ k" 

In the reaction A + 2B <= C + D (i.e., A + B + B^C + D), 
the rate of reaction between A and B is proportional to the con- 

46 



EQUILIBRIUM CONSTANTS 47 

cent ration of A and to the square of the concentration of B. Hence, 
at equilibrium, 

[ApP = K 

More generally, in the reaction wA + #B + <=* yC + zD + , 
the equilibrium constant is expressed as follows: 

IAMB? = K 

For the value K to be approximately a true constant, con- 
centrations must be relatively small (see Sec. 32). Furthermore, 
although in computations involving chemical equilibria, concentra- 
tions are better expressed in terms of the number of moles of sub- 
stance per kilogram of solvent, for dilute solutions this is practically 
the same as the number of moles of substance per liter of solution; 
and as the latter is consistent with the definition of a molar solu- 
tion as used in analytical computations, it will be employed here 
in formulating mass-action expressions as well. In mass-action 
expressions molar concentrations will be represented as above by 
enclosing in brackets the symbol of the element, compound, or 
radical in question. 

In general, if a solid substance is involved in a chemical equi- 
librium, its concentration is not included in the formulation of 
the mass-action constant, since the concentration of the solid is 
itself essentially a constant. The same is true of water in an 
equilibrium involving dilute aqueous solutions. Thus, the mass- 
action constant for the dissociation equilibrium 

NH 4 OH ^ NH 8 + H 2 O 
is simply 

[NH 3 ] K 
[NH 4 OH] 

27. Ion Product Constant of Water. Water dissociates slightly 
into hydrogen ions * and hydroxyl ions as follows: 

H 2 <=> H+ + OH- 

1 Experiments indicate that the hydrogen ion is hydrated. It is therefore 
often expressed as BUG*. This ion is called the hydronium ion and is formed 
by the union of a proton with a molecule of the solvent. The use of this symbol 
complicates equations and offers no particular advantages in analytical 
computations. 



48 CALCULATIONS OF ANALYTICAL CHEMISTRY 

The mass-action expression for this dissociation is simply 

[H+][OH-] = Ku, 

since the concentration of undissociated H 2 in dilute aqueous 
solutions is essentially a constant and, as stated above, is omitted 
from mass-action expressions. In any aqueous solution, there- 
fore, the product of the molar hydrogen-ion concentration and 
the molar hydroxyl-ion concentration is a constant at a given 
temperature. This constant is called the ion product constant of 
water and at 25C. has a value of 1.0 X 1Q- 14 . 

[II+][OH-] - Ku, = 1.0 X 10~ 14 (at 25C.) 
(= 1.2 X 10~ 15 at 0C.) 
(= 5.8 X 10- 13 at 100C.) 

In pure water the hydrogen-ion and the hydroxyl-ion concentra- 
tions are equal; at 25C. each has a value of 1.0 X 10~ 7 molar. 

28. pH Value. It is often convenient to express hydrogen-ion 
concentrations in terms of the pH value. The pH value as used 
in analytical chemistry is simply the common logarithm of the 
reciprocal of the molar hydrogen-ion concentration. 

pH = log ^ = - log [H+] = colog [H+] 

Similarly the pOH value, although less often used, is the loga- 
rithm of the reciprocal of the hydroxyl-ion concentration. The 
pH value of pure water at 25C. is 7. The pH value of acid solu- 
tions is less than 7; the pH value of alkaline solutions is greater 
than 7. In general at 25C., 

pH + pOH = 14 

EXAMPLE I. What is the pll value and what is the hydroxyl- 
ion concentration of a solution that is M/ 1,000 in HC1 (effective 
ionization = 100 per cent)? 

SOLUTION: 

[H+] = 0.001 = 1 X 10- 3 

pH = log 1 x 1Q _ 3 = 3. Ans. 

pOH = 14 - 3 = 11 
[OH-] = 1 X 10-". Ans. 



EQUILIBRIUM CONSTANTS 49 

EXAMPLE II. The hydrogen-ion concentration in a certain 
dilute solution of sulfuric acid is 2.0 X 10~ 5 . What is the pH 
value? What is the pOH value? 
SOLUTION: 

(5 x 



log 10 4 + log 5 = 4 + 0.70 = 4.70. Ana. 
pOH = 14 - 4.70 = 9.30. Ana. 

EXAMPLE III. The pH value of a certain solution is 5.92. 
What is the pOH value, the hydrogen-ion concentration, and the 
hydroxyl-ion concentration? 
SOLUTION: 

pH + pOH = 14 

pOH = 14 - 5.92 = 8.08. Ans. 
[H+] = 10- 5 - 92 = 10+- 08 X 10- 6 = 1.20 X 10- 6 . Are*. 
[H+][OH-] = 1.0 X 10- 14 



i o 
[OH ~ ] = 1 20 X 10- = 8 ' 3 X 

Problems 
I (Temperatures are 25*C.) 

87. (a) What is the pH value of a solution in which the hydrogen-ion 
concentration is 2.8 X 10~ 3 ? Is the solution acid or alkaline? (6) What is 
the hydrogen-ion concentration of a solution with a pOH value of 4.17? Is 
the solution acid or alkaline? 

Ans. (a) 2.55, acid. (6) 1.5 X lO" 10 , alkaline. 

88. What is the pH value of 0.010 molar IIC1 (100 per cent ionized)? Of 
0.30 molar NaOH (90 per cent effective ionization)? Of a solution of HC1 in 
which the hydrogen-ion concentration is 8 molar? 

Ans. 2.0. 13.43. - 0.90. 

89. What is the hydrogen-ion concentration of a solution in which pH = 
- 0.55? 

Ans. 3.6 molar. 

90. (a) Given pH = 10.46. Calculate [H+], [OH~], and pOH. (6) Given 
[Oil-] = 5.6 X 10- 2 . Calculate [H+], pH, and pOH. 

Ans. (a) 3.5 X 10-", 2.9 X 10~ 4 , 3.54. (6) 1.8 X 10~ 13 , 12.75, 1.25. 



91. (a) What is the pOH value of a solution in which the hydrogenion 
concentration ^s 5.3 X 10~ 4 ? Is the solution acid or alkaline? (6) What is the 



50 CALCULATIONS OF ANALYTICAL CHEMISTRY 

hydroxyl-ion concentration of a solution in which the pH value is 9.27? Is 
the solution acid or alkaline? 

92. What is the pH value of 0.050 molar IINO 3 (100 per cent ionized)? 
Of 0.80 molar KOH (effective ionization = 85 per cent)? Of a solution of HC1 
in which the hydrogen-ion concentration is 5.0 molar? 

93. What is the hydroxyl-ion concentration of a solution in which pOH = 
- 0.27? 

94. (a) Given pOH = 5.80. Calculate [H+], [Oil-], and pH. (6) Given 
[H]+ = 3.1 X 10~ 9 . Calculate IOH~], pll, arid pOH. 

29. Ionization Constant. The law of mass action can be applied 
to the equilibrium in dilute solution between the molecules of a 
weak acid or weak base and its ions. Thus acetic acid, HC 2 H 3 O 2 , 
is partially ionized in solution as follows: 

HC 2 H 3 2 <=> H+ + C 2 H 3 O 2 - 
Therefore, 

[H+][C 2 H 3 Q 2 ~] _ v 
[HC 2 H 3 2 ] ~ Vnc > H >* 

That is, in a solution containing acetic acid, the total molar con- 
centration of hydrogen ions (from whatever source) multiplied by 
the total molar concentration of acetate ions (from whatever 
source) divided by the molar concentration of un-ionized acetic 
acid, is a constant at a given temperature. This value is called 
the ionization constant of acetic acid. Its value at 25C is 1.86 X 

io- 5 . 

A similar mass-action expression can be set up for the ionization 
of a solution of ammonia in water. Here ammonium hydroxide 
molecules are presumably formed which, in turn, dissociate par- 
tially into ammonium ions and hydroxyl ions. 

NH 4 OH <=> NH 4 + + OH- 
The ionization constant is therefore usually written 

[NH 4 +][OH-1 _ 
[NH 4 OH] " KNH < OH 

Since ammonium hydroxide molecules (if they exist at all) are in 
equilibrium with ammonia and water, the equilibrium is more 
generally expressed as 

H 2 ?= (NH 4 OH) * NH 4 + + OH" 



EQUILIBRIUM CONSTANTS 51 

and the ionization constant can be written 

[NH 4 +][OH-] _ 



In either case the total concentration of NH 3 , either dissolved as 
such or combined as NII 4 OH, is used in the denominator of the 
fraction, so the numerical value of the constant is the same in 
the two cases. At 25C. it is 1.75 X 10~ 5 . 

The ionization constants of a few weak acids and bases are 
given in the Appendix. 

EXAMPLE I. What is the ionization constant of acetic acid at 
a certain temperature if in tenth-molar solution it is 1.3 per cent 
ionized? 

HC 2 H 3 2 <= H+ + C 2 H 3 O 2 - 

SOLUTION: If 0.10 mole of HC 2 H 3 O 2 were completely ionized, 
it would give 0.10 mole (or 0.10 gram-ion) of H+ and 0.10 mole 
(or 0.10 gram-ion) of C 2 H 3 O 2 ~. Being only 1.3 per cent ionized, 
it gives 0.10X0.013 = 0.0013 mole of 11+ and 0.0013 mole of 
C 2 H 3 O 2 ~, leaving 0.0987 mole of undissociated HC 2 H 3 2 mole- 
cules. The molar concentrations are therefore as follows: 

[11+] = 0.0013 

[C 2 H 3 2 -] = 0.0013 

[HC 2 H 3 2 ] = 0.0987 

Substituting these in the above expression for the ionization con- 
stant of acetic acid, we get 

(0.001 3) (0.0013) _ 
0.0987 ~ 

K = 1.7 X 10- 5 . Ans. 

EXAMPLE II. At 25C. the ionization constant of acetic acid 
is 1.86 X 10~ 5 . What is the molar concentration of hydrogen ions 
in a 0.20 molar solution of acetic acid at 25C.? 
SOLUTION: 

Let x = molar concentration of H+ 
Then 

x = molar concentration of C 2 H 3 0<r 
and 

0.20 x = molar concentration of undissociated HC 2 H 3 O? 



52 CALCULATIONS OF ANALYTICAL CHEMISTRY 



(0.20 - x) 
Solving, 

x = 1.9 X 10~~ 3 mole per liter. Ans. 

Numerical values in mass-action expressions need not be ex- 
pressed to more than two or, at the most, three significant figures. 
Therefore simplifying assumptions can often be made. In the 
above fractional equation, the value of x is so small compared 
with the value 0.20 from which it is subtracted, that it is well 
within the limit of precision to write 

(*)(*) 



0.20 



= 1.8GX 10~ 5 



and thus avoid solving a quadratic equation. 1 

30. Common Ion Effect. Buffered Solution. Suppose into a 
dilute solution of acetic acid is dissolved a considerable quantity 
of sodium acetate, i.e., a highly ionized salt of acetic acid. The 
total acetate-ion concentration is greatly increased, but since the 

ri . . [H + ][C 2 H 3 O 2 -] , , 

equilibrium constant ^-TFT^-TT/A == K H c,H 3 o 2 must be main- 



tained, the greater part of the hydrogen ions present must unite 
with acetate ions to form more of the undissociated acetic acid 
molecules. In other words, the equilibrium reaction HC 2 H 3 2 <=* 
H+ + C 2 H 3 2 ~ must go to the left to a degree sufficient to rees- 
tablish the numerical value of the constant. The solution there- 
fore becomes much less acidic, with a hydrogen-ion concentra- 
tion only slightly greater than that of pure water. 

A similar case is one in which an ammonium salt is added to a 
solution of ammonium hydroxide. The common ion, NH4 4 ", re- 
presses the ionization NH 3 + H 2 O <F (NH 4 OH) < NH 4 + + OH- to 
a very great extent, for in order to maintain the equilibrium con- 

. , [NH 4 +][OH-] ^ . , . . ^. , , 

stant i f ^" 1 - = K, the hydroxyl-ion concentration must be 



1 In cases where such simplifying assumptions cannot be made, the formula 
for solving the general quadratic equation 



bx + c = 0| is x = - 

2a 



EQUILIBRIUM CONSTANTS 53 

greatly decreased. The resulting solution is only slightly more 
basic than pure water. 

In each of the above two cases the solution is said to be buffered 
by the common ion added. The acetic acid-sodium acetate com- 
bination, for example, has a low hydrogen-ion concentration which 
is little affected even by the addition of small amounts of a strong 
acid, for the additional hydrogen ions merely unite with the acetate 
ions (still present in excess) to give more acetic acid, the ioniza- 
tion of which is repressed by the acetate. Similarly, the basicity 
of a buffered ammonium hydroxide solution is not much affected 
even by the addition of small amounts of a strong base like sodium 
hydroxide. 

Buffered solutions are much used both in qualitative and quanti- 
tative analysis to effect certain separations of elements where a 
carefully controlled hydrogen-ion or hydroxyl-ion concentration 
is essential. Problems covering some of these separations are in- 
cluded in this book. 

EXAMPLE. What is the hydrogen-ion concentration in 500 ml. 
of a 0.10 molar solution of acetic acid at 25C. if the solution con- 
tains an additional 2.0 grams of acetate ions added in the form 
of sodium acetate? (K H c 2 H,o 2 = 1.86 X JO" 5 .) 
SOLUTION : 

2 grams C2H 3 02~~ per 500 ml. = 0.068 mole per liter 

Let x = concentration of H + ions 
Then 

x + 0.068 = concentration of C2H 3 O 2 ~ ions 
and 

0.1 x = concentration of un-ionized 
Therefore, 



or approximately, 

W|JP_ 1.86X10-' 

(since x is small) 
Solving, 

x = 2.6 X 10~ 5 mole per liter. Ans. 

31. lonization of Polybasic Acids. Polybasic acids like 
H 2 C0 3 , H 3 P0 4 , etc., ionize in steps, and a mass-action expression 



54 CALCULATIONS OF ANALYTICAL CHEMISTRY 

can be written for each step. Thus, H 2 S ionizes to form H+ and 
HS~, in which case the ionization constant is 

__ T^ _ Q i v in -8 
- Kl ~ 9>1 X 10 

The HS~ ions are further ionized into H+ and S = , in which case 



Multiplying the two equations one by the other gives 



[H 2 S] -- 

A saturated solution of H 2 S is about 0.10 molar, and [H 2 S] = 
0.10. Therefore in cases where metallic elements are precipitated 
by saturating their solutions with H 2 S, [H+] 2 [S=] = 1.1 X 10~ 23 . 

It is seen that the primary ionization of H 2 S is much greater 
than the secondary ionization and that the ionization cannot be 
correctly expressed by the equation H 2 S <=^ 2H+ + S~. The con- 
centration of H+ ions in a solution of H 2 S is not twice that of the 
S^ ions. The primary ionization of any polybasic acid is much 
greater than the secondary ionization. 

EXAMPLE I. What is the approximate hydrogen-ion concen- 
tration in a solution of hydrogen sulfide which is 0.07 molar in H 2 S? 

rTT4-i 2 ra=i 

SOLUTION: In solving this problem, the expression 



1.1 X 10~ 22 cannot be used since neither [H+] nor [S^] is known 
and there is no simple relation between them. On the other hand, 
although H 2 S is ionized in two steps, the first ionization is so much 
greater than the second ionization that, for the purpose of obtain- 
ing an approximate answer, the latter may be considered negligible. 
In other words, practically all of the hydrogen ions may be con- 
sidered to come from the ionization of H 2 S into H+ and IIS"". 
Therefore [H+] and [HS~] are practically equal in value and 

[H+KHS-] 1Q _ 8 

[H 2 S] J>1 X U 
or , 

- 9.1 X 10- 



x = 8 X 10~ 5 mole per liter. Ans. 



EQUILIBRIUM CONSTANTS 55 

EXAMPLE II. What is the concentration of S 8 * ions in 200 ml. 
of a solution that is 0.05 molar in H 2 S and that by the addition 
of HC1 contains a total of 0.12 equivalent of H+ ions? 

SOLUTION: 

Let x = molar concentration of S = ions 

0.05 x = approx. 0.05 = concentration of undissociated 
0.12 X 5 = 0.6 = concentration of H+ ions 



(0.05) -- 

x = 1.5 X 10~ 23 mole per liter. Ans. 

32. Activity and Activity Coefficients. In analytical chemistry 
mass-action calculations are usually applied to equilibria involving 
electrolytes in solution. As solutions of electrolytes are made pro- 
gressively more concentrated, the quantitative effect on such prop- 
erties as conductivity and freezing-point lowering becomes pro- 
gressively less than that calculated solely from the net change in 
molar concentration. This is likewise true of mass-action equi- 
libria. This phenomenon was formerly explained by assuming 
that electrolytes are less completely ionized in more concentrated 
solutions; that the degree of ionization approaches 100 per cent 
only as dilution approaches infinity. A more satisfactory explana- 
tion is based on the assumption that most salts and the so-called 
strong acids and bases are practically completely ionized in all 
aqueous solutions but that the effective concentration, or activity, 
of the ions is decreased because of forces of attraction between 
the positive and negative ions. These forces become less at higher 
dilutions since the ions are farther apart. 

In mass-action expressions, therefore, activities or effective con- 
centrations, rather than molar concentrations should be used for ac- 
curate results. The activity (a) of an ion or molecule can be found by 
multiplying its molar concentration (c) by an activity coefficient (/). 

a =/c 

An activity coefficient is therefore a factor which converts a molar 
concentration to a value which expresses quantitatively the true 
mass-action effect. Thus, the ionization constant of acetic acid 
is correctly expressed as 

/i[H+]X/,[C,H,Qr] = 



56 CALCULATIONS OF ANALYTICAL CHEMISTRY 

where /i, / 2 , and / 3 are the activity coefficients of the hydrogen 
ion, the acetate ion, and the acetic acid molecule, respectively. 

The numerical value of an activity coefficient is not always 
easy to determine, since it depends on many factors such as tem- 
perature, the number of charges on the ion, and certain electrical 
properties of the solution. In the case of relatively dilute solu- 
tions (e.g., 0.01 formal or less) and particularly for univalent ions, 
activity coefficients are not far from unity, and so no great error 
is introduced when molar concentrations are used in place of ac- 
tivities. Since concentrations in most analytical operations are 
relatively low and since a high degree of precision is seldom re- 
quired in analytical computations involving mass-action constants, 
activity coefficients can be omitted without much error. They are 
therefore not included in the calculations in this book. 

33. Dissociation Constants of Complex Ions. A complex ion, 
by definition, is one that is in equilibrium with its constituents. 
These constituents are ordinarily a simple positive ion and either 
a neutral molecule or a negative ion. The mass-action principle 
can be applied to dilute solutions of such ions. Thus the copper 
ammino (or copper ammonio) ion, Cu(NH 3 )4" M ", ionizes slightly 

as follows: 

Cu(NH 3 ) 4 ++ * Cu++ + 4NH 3 

Its dissociation constant is therefore expressed thus, 

[Cu++][NH 3 ]* _ , u) 

[Cu(NH 3 ) 4 ++ ] ( 4.0 xiu ; 

This means that in a dilute solution containing the complex ion, 
the total molar concentration of the simple cupric ions present, 
multiplied by the fourth power of the total molar concentration 
of ammonia (NH 3 + NH 4 OH), divided by the molar concentra- 
tion of the undissociated complex ion, is a constant at a given 
temperature. 

Complex ions of this type frequently encountered in analytical 
chemistry are Ag(NH 3 ) 2 +, Cu(NH 3 ) 4 ++, Cd(NH 3 ) 4 ++, Ni(NH 3 ) 4 ++ 
CoCNHah^, and Zn(NH 3 ) 4 ++. 

Important cyanide complexes include Fe(CN) 6 s , Fe(CN) 6 s , 
Ag(CN)r, Cd(CN) 4 =, Cu(CN)a- Hg(CN)r, Co(CN) 6 ^, and 
Ni(CN)4". Halide complexes like SnClr and HgI 4 ~ and oxalate 
complexes like Mg(C 2 4 )<r are also commoa 



EQUILIBRIUM CONSTANTS 57 

EXAMPLE I. What is the molar concentration of mercuric ions 
and of cyanide ions in a tenth-molar solution of K2Hg(CN)4? 
(Dissociation constant of Hg(CN) 4 = - 4.0 X 1Q- 42 .) 

SOLUTION: 

[Hg++][CN-f 

[Hg(CN)r] - 4 - oxl 

Let x = concentration of Hg ++ in the dissociation: 

Hg(CN) 4 = <=* Hg++ + 4CN- 

Then 

4x = concentration of CN~ 

0.10 x 0.10 (approximately) = concentration of Hg(CN)4~ 
= 4.0 X 10~ 42 



x = 1.1 X 10~ 9 molar Hg++. Ans. 
4x = 4.4 X 10~ 9 molar CN~. Ans. 

EXAMPLE II. What is the dissociation constant of Ag(NHs)2 + 
if a solution of 0.20 formula weights of AgCl, in sufficient excess 
NI^OH to give a total ammonia concentration of 2.0 molar and 
a total volume of one liter, has a silver-ion concentration of only 
0.00037 mg. per liter? 
SOLUTION : 

[Ag+][NH 3 ? _ 
[Ag(NH 3 ) 2 +] 

[Ag+] = a 003 1 n < y 10 ~ 8 = 3.4 X 10- 9 mole per liter 



[NH,] = 2.0 

[Ag(NH 3 ) 2 + ] = 0.20 (approximately) 
(3.4xlO- 9 )(2.0) 2 _ ir 

0.20 

= 6.8 X 10~ 8 . Ans. 

Problems 

(See Appendix for ionization constants and dissociation constants. Tem- 
peratures are 25C. unless otherwise specified.) 

96. A certain organic acid has one replaceable hydrogen and in 0.010 molar 
aqueous solution is 0.18 per cent ionized. What is the ionization constant of 
the acid? 

Ans. 3.2 X 10-. 



58 CALCULATIONS OF ANALYTICAL CHEMISTRY 

96. Lactic acid is a monobasic acid with an ionization constant of 1.6 X 
10~ 4 . What is the lactate-ion concentration in a 0.50 N solution of the acid? 

Arts. 8.9 X 10~ 3 molar. 

97. What is the molar concentration of each of the three constituents of 
acetic acid in 0.050 N solution? 

Ans. H+ = 0.00096, C 2 H 3 Or = 0.00096, HC 2 H 3 O 2 = 0.049. 

98. What is the concentration of a solution of NII 4 OII if it is 3.0 per cent 
ionized? If it is 0.50 per cent ionized? 

Ans. 0.020 molar. 0.72 molar. 

99. Formic acid is a monobasic acid that is 3.2 per cent ionized in 0.20 molar 
solution. What is the ionization constant of formic acid, and what is its 
percentage ionization in 0.050 molar solution? 

Ans. 2.1 X 10~ 4 , 6.4 per cent. 

100. What is the hydrogen-ion concentration in a 0.10 normal solution of 
acetic acid containing sufficient dissolved sodium acetate to give a total acetate- 
ion concentration of 0.85 mole per liter? 

Ans. 2.2 X 10-* molar. 

101. What is the hydrogen-ion concentration and the pOH value of a 0.010 
molar solution of hypochlorous acid at 25C.? 

Ana. 2 X 10~ 5 molar, 9.30. 

102. What is the pH value of a 0.30 normal solution of NH 4 OII? What 
is the pH value of a 0.30 normal solution of NH 4 O1I containing sufficient dis- 
solved NH4C1 to give an ammonium-ion concentration of 1.2 moles per liter? 

Ans. 11.36. 8.63. 

103. Approximately how many grams of acetate ions should be dissolved 
in a liter of 0.10 M acetic acid in order to cut down the hydrogen-ion con- 
centration one hundredfold? 

Ans. 8 grams. 

104. To what volume should 100 ml. of any weak 0.30 molar monobasic 
acid or mono-acidic base be diluted in order to triple its percentage ionization? 

<Ans. 900 ml. 

/%t)6, What is the approximate concentration of sulfide ions and of hydro 
mlfide ions (HS") in a 0.070 molar solution of hydrogen sulfide? (Hint: As- 
sume that practically all the hydrogen ions come from the primary ionization,) 
Ans. 1.2 X 10~ 1B molar, 8 X 10~ 6 molar. 

106. Calculate the concentration of sulfide ions in a solution which is 
0.080 molar in H 2 S and contains sufficient HC1 to give a pH value of 3.40. 

Ans. 5.5 X 10~ 17 molar. 

107. What is the approximate molar concentration of silver ions and of 
cyanide ions in a tenth-molar solution of KAg(CN)2? [Dissociation constant 
of Ag(CN) 2 - = 1.0 X 10- 21 ]. 

Ans. 3 X 10~ 8 molar, 6 X 10~ 8 molar. 



EQUILIBRIUM CONSTANTS 59 

108. What are the approximate molar concentrations of Na + , Cd + , CN~, 
and Cd(CN) 4 - in a solution made by dissolving 0.020 F.W. of Na 2 Cd(CN) 4 in 
water and diluting to one liter? 

Aru. [Na+] = 0.040 molar, [Cd ++ ] = 6.4 X 10" 5 molar, [ON"] = 2.5 X 
10~ 4 molar, [Cd(CN) 4 ~] = 0.020 molar. 

109. If 100 milligrams of AgCl are dissolved in excess ammonium hydroxide 
to give a volume of 500 ml. of solution, and the total concentration of ammonia 
is 0.30 molar, what is the silver-ioii concentration? Dissociation constant of 
Ag(NH 3 ) 2 + = 6.8 X 10~ 8 . 

JLns. 1.06 X 10~ 9 molar. 

110. What is the concentration of Cd++ ions in a solution 0.040 molar in 
Cd(NH 3 ) 4 ++ and 1.5 molar in NH 3 ? 

Ans. 2.0 X 10~ 9 molar. 



111. A certain organic amine acts as a mono-acidic base in aqueous solu- 
tion. A 0.05 molar solution is found to give a hydroxyl-ion concentration of 
7.5 X 10~ B molar. What is the ioiiization constant of the base? What is its 
pll value? 

112. Lactic acid (HC 3 H 6 O2) is a monobasic acid with an ionization constant 
at 25C. of 1.6 X 10" 4 . In a tenth-molar solution how many grains of lactic 
acid are present in the un-ionized form? 

113. What is the molar concentration of the three constituents of benzoic 

acid in 0.080 M solution at 25C.? What is the pH value? 

\ 

114. What molarity acetic acid is 2.0 per cent ionized at 25C.? 

115. Ethylamine is a derivative of ammonia and in aqueous solution is basic 
like ammonia. At a certain temperature ethylamine in 0.30 molar solution 
gives a hydroxyl-ion concentration of 1.3 X 10~ 2 molar. What is the ioniza- 
tion constant of ethylamine at that temperature, and what is its percentage 
of ionization in 0.20 molar solution? 

116. Calculate the cyanide-ion concentration and the percentage ionization 
of a 0.030 molar solution of hydrocyanic acid. 

Xll7. Calculate the hydrogen-ion concentration of a solution at 25C. con- 
taining 25 ml. of 4 N acetic acid in a total volume of 1,200 ml. Calculate the 
hydrogen-ion concentration in the same solution after adding 15 grams of 
sodium acetate (assuming the effective ionization of the salt to be 85 per cent). 
What is the pH value in each case? 

118. To what volume should 50 ml. of any weak 0.20 molar monobasic 
acid be diluted in order to double its percentage ionization? 

119. In a 0.20 molar solution of ammonium hydroxide, what percentage of 
the base is un-ionized? What is its pH value? 

120. Approximately how many grams of NH 4 + ions should be dissolved 
into a liter of 0.20 M NH 4 OH in order to cut down the concentration of hy- 
droxyl ions to one-fiftieth its previous value? 



60 CALCULATIONS OF ANALYTICAL CHEMISTRY 

121. What is the pH value of a 0.25 normal solution of acetic acid? What 
is the pH value of a 0.25 normal solution of acetic acid containing sufficient 
dissolved sodium acetate to give an acetate-ion concentration of 2.0 moles per 
liter? 

122. What are the approximate concentrations of HCOs" and of COs*" in a 
0.0010 molar solution of carbonic acid? (Hint: Assume that practically all 
of the hydrogen ions come from the primary ionization of the acid.) 

123. What is the sulfide-ion concentration of a solution 0.090 molar in E^S 
and containing sufficient HC1 to give a pH value of 4.50? 

124. What are the approximate molar concentrations of K + , Hg++, HgI 4 " 
and I" in a solution made by dissolving 0.010 F.W. of K 2 HgI 4 in water and 
diluting to one liter? (Dissociation constant of HgI 4 = = 5.0 X 10~ 31 ). 

125. What is the approximate molar concentration of cyanide ion in a 
solution 0.010 molar in K 4 Fe(CN) 6 ? 

126. If 50 milligrams of AgCl are dissolved in excess NH 4 OH (AgCl + 
2NH 4 OH -> Ag(NH 8 ) 2 + + Cl~ + 2H 2 O) and the resulting solution is 0.50 
formal in NH 3 and has a volume of 500 ml., what is the concentration of Ag + 
ions in formula weights per liter? 

127. What is the molar cyanide-ion concentration of an aqueous solution 
containing 0.020 F.W. of K 2 Ni(CN) 4 per 500 ml.? What is the concentration 
of Ni++ in such a solution if sufficient additional KCN is present to give a total 
cyanide concentration of 0.10 molar? 

^SL28. If 0.10 F.W. of Hg(NO 3 ) 2 is treated with excess Na 2 S solution in the 
presence of NaOH, the precipitate of IlgS that first forms dissolves to give 
HgS 2 ~ ions. If the dissociation constant of HgS 2 ~ is 2 X 10~ 56 and the sulfide- 
ion concentration of the solution is 2.0 molar, what is the concentration 
of Hg++? 

34. Solubility Product. A very important equilibrium constant 
applies to a saturated solution of a slightly soluble, completely 
ionized salt. Most of the precipitates encountered in analytical 
chemistry belong to this category. 

Consider the simple case of a saturated solution of silver chloride 
in equilibrium with some of the undissolved salt. What little silver 
chloride, is in solution is completely ionized, and the equilibrium 
can be written 

AgCl(solid) <=> Ag+ + Cl- 



The mass-action equilibrium constant is expressed simply as 
or more accurately as 



-] = K AgCl 



EQUILIBRIUM CONSTANTS 61 

where /i and / 2 are the respective activity coefficients of the two 
ions (see Sec. 32). These coefficients are only slightly less than 
1.00 in value. 

This constant, applying as it does to a saturated solution of a 
slightly soluble salt, is called a solubility product (K S . P .). The 
numerical value of the solubility product of silver chloride at 
25C. is 1.0 X 10" 10 . This means that in a solution saturated 
with silver chloride at this temperature the total molar concen- 
tration of silver ions in the solution multiplied by the total molar 
concentration of chloride ions equals 1.0X10~ 10 . Conversely, 
when the product of the total concentration of silver ions and 
the total concentration of chloride ions in any solution exceeds 
this value, a precipitate of silver chloride is obtained under con- 
ditions of stable equilibrium. 

Lead chloride ionizes as follows: 

PbCl 2 * Pb++ + 2C1- 
Its solubility product is therefore 

[Pb++][Cl-]* = K PbC1 , 

Here the square of the total chloride-ion concentration must be 
used. In terms of activities, the solubility product is 



In most mass-action calculations two significant figures are all 
that are warranted by the precision of the data and of the con- 
stant itself. The precision is much less in calculations involving 
the solubilities and solubility products of the more insoluble hy- 
droxides and sulfides. These values are usually known only very 
approximately, for the composition of a precipitate of this type 
may be quite variable. 

EXAMPLE I. What is the solubility product of Ag 3 P0 4 if the 
solubility of the salt is 6.5 X 10~ 3 gram per liter? 

SOLUTION: 6.5 X 10~ 3 gram per liter = (6.5 X 10- 3 )/418.7 = 
1 .6 X 10~~ 6 mole per liter. The salt is 100 per cent ionized as follows : 



Therefore, 

[Ag+] = 3 X 1.6 X 10~ 5 
[PO 4 S ] = 1.6 X 10~ 5 
(3 X 1.6 X 10" 6 ) 3 (1.6 X 10- 5 ) = 1.8 X 10~ 18 . Ans. 



62 CALCULATIONS OF ANALYTICAL CHEMISTRY 

EXAMPLE II. The solubility product of CaF 2 is 3.2 X 10~ u . 
How many grams of Ca ++ are present in 500 ml. of a saturated 
solution of CaF 2 ? How many grams of CaCU can be dissolved 
in 500 ml. of a solution containing 9.5 grams of fluoride ions? 
SOLUTION: 

Let x = molar concentration of Ca" 1 " 4 " 
Then 

2x = molar concentration of F~ 

(z)(2z) 2 = 3.2x 10- u 

x = 2.0 X 10- 4 mole per liter 

2.0 X 10- 4 X 40 X 5 = 0.0040 gram Ca++ per 500 ml. Ans. 

Zi 

9.5 grams F~ per 500 ml. = 1 mole F~ per liter 
(z)(l) 2 = 3.2X 10- 11 

x = 3.2 X 10-" mole Ca ++ per liter 
3.2 X 10~ n X 111 = 2.9 X 10- 9 gram CaCl 2 per liter 

= 1.45 X 10- 9 gram CaCl 2 per 500 ml. Ans. 

EXAMPLE III. What is the hydroxyl-ion concentration in a 
solution of sodium hydroxide having a pH value of 11.6? How 
many grams of magnesium could remain dissolved in 500 ml. of 
such a solution [solubility product of Mg(OH) 2 = 3.4 X 10~ u ]? 
SOLUTION: 



[H+] = 10- 11 - 6 = 10 ( ~ 12 + - 4) or 

= 2.52 X 10- 12 (since antilog 0.4 = 2.52) 

i o v 1ft- 14 

[OH " ] = 2.52 X10-" = 3 ' 98 X 10 ~ 3 ' AnS - 
pOH = 14 - 11.6 = 2.4 

[OH-] = 10- 2 - 4 = 10- 6 X 10~ 3 = 3.98 X 10- 3 . Ans. 
[Mg++][OH-] 2 = 3.4 X 10- 11 
4. V 10- 11 

^ = 2A X 10 ~ 6 m le per liter = 



2.55 X 10~ 5 gram per 500 ml. Ans. 

36. Fractional Precipitation. Ordinarily when a precipitating 
agent is added slowly to a solution containing two ions capable 
of being precipitated by the agent, the substance with the lesser 
solubility will precipitate first. The point at which the second 
substance will precipitate can be determined from the solubility 
products of the two precipitates. 



EQUILIBRIUM CONSTANTS 63 

Suppose to a solution 0.10 molar in Ba++ and 0.10 molar in 
is added gradually and in very minute quantities a solution 
of Na 2 SO4. Insoluble BaSO 4 (solubility product, K Ba so 4 = 1.1 X 
10~ 10 ) precipitates first, then SrSO 4 (K SrS04 = 2.8 X 10~ 7 ) begins 
to precipitate. The ratio of the two solubility products is as follows : 

[Ba++][SOrl _ 1.1 X 10- 10 
[Sr++][SOr] 2.8 X 10~ 7 
Therefore, 

rRfl++i 

0J - 0.00039 

At the point where SrSO4 just begins to precipitate (and the con- 
centration of Sr++ is still 0.10 M) the barium-ion concentration 
will have been reduced to 0.000039 molar, since 

[Bar"-] 



0.10 



= 0.00039 



Separation of the two cations is therefore nearly complete at this 
point. 

In qualitative analysis the preparation of a solution of a water- 
insoluble salt for the anion tests is usually made by metathesis 
of the solid with a solution of Na 2 C(V The extent of metathesis 
can be determined, roughly at least, from the solubility product 
of the original salt and that of the insoluble compound of the 
metal formed by the metathesis. 

EXAMPLE. If lead iodide (PbI 2 ) is boiled with 2.0 M Na 2 C0 3 
solution, the insoluble lead iodide is converted to the more in- 
soluble lead carbonate (PbI 2 + COs^ 21" + PbC0 3 ). Assuming 
that sufficient PbI 2 is present to give equilibrium conditions be- 
tween the two insoluble substances and that the solubility products 
of PbI 2 and PbCOs at the temperature of the solution are 2.4 X 
10~ 8 and 5.6 X 10~ 14 , respectively, what would be the concentra- 
tion of iodide ion in the resulting solution? 
SOLUTION: 

2.4 X 10-* 
5.6 X 



. 

[I~] = 930 moles per liter. Ans. 



64 CALCULATIONS OF ANALYTICAL CHEMISTRY 

This concentration is of course impossible to attain, not only be- 
cause of the limited solubility of the Nal formed, but also because 
the Pbl2 is completely metathesized before the equilibrium condi- 
tion is reached. The result merely shows that PbI 2 is readily 
and completely metathesized by Na2C03 solution. 

Problems 

(Temperatures are 25C. unless otherwise specified. A table of solubility 
products is given in the Appendix.) 

129. A saturated solution of barium fluoride, BaF 2 , is 7.5 X 10~ 3 molar. 
What is the solubility product of barium fluoride? 

Am. 1.7 X Wr+. 

130. If 0.11 mg. of silver bromide dissolves in one liter of water, what is 
the solubility product of silver bromide? 

Am. 3.5 X 10~ 13 

181. If the solubility product of lead phosphate, Pb 8 (PO 4 )2, is 1.5 X 10" 32 , 
how many milligrams will dissolve in 500 ml. of water? How many milligrams 
of Pb"*~+ can remain dissolved in 500 ml. of a solution that is 0.20 molar in 
PO 4 " ions? 

Am. 0.070 mg. 7.5 X lO" 6 mg. 

132. From the solubility product of PbI 2 calculate how many grams of 
lead ions and of iodide ions are contained in each milliliter of a saturated solu- 
tion of lead iodide? 

Am. 3.7 X 10~ 4 , 4.6 X 10~ 4 gram. 

133. What are the solubility products of PbC 2 O 4 and of Pb(IO 8 ) 2 if the 
solubilities are 1.7 X 10~ 3 gram per liter and 1.6 X 10" 2 gram per liter, 
respectively? 

Am. 3.3 X 10- 11 , 9.8 X 10~ 14 . 

134. If the solubility products of AgBrO 3 and of Ag 2 Cr 2 O 7 are 5.0 X 10~ 5 
and 2.7 X 10~ n , respectively, what are the solubilities of the two salts in 
milligrams per 100 ml.? 

Am. 167 mg., 8.2 mg. 

136. A saturated solution of K 2 PtCle contains 11 mg. of the salt in each 
milliliter. What is the solubility product of the salt? How many milligrams 
of Pt can remain dissolved (as PtCle") in each milliliter of a solution that 
contains 3.9 grams of K+ per liter? 

Am. 4.6 X 10~ 6 . 0.90 mg. 

136. Mercurous bromide, Hg 2 Br 2 , dissociates into Hg 2 " f " h and 2Br~~. Its 
solubility at 25C. is 0.039 mg. per liter. What is its solubility product at 
that temperature? 

Am. 1.4 X 10-* 




EQUILIBRIUM CONSTANTS 65 

137. The normality of a saturated solution of Ag 2 COs is 2.2 X 10~ 4 N. 
What is the solubility product of A&CO*? 

Ans. 5.3 X 10- 12 . 

138. If A moles of AgsP0 4 dissolve in 500 ml. of water, express the solu- 
bility product of Ag3PO 4 in terms of A, and the normality of a saturated solu- 
tion of Ag 3 PO4 in terms of A. 

Ans. 432A 4 , 6A. 

139. If the solubility product of Ca8(PO 4 ) 2 is A, express in terms of A the 
solubility of Cas(PO 4 ) 2 in moles per liter. Also express in terms of A the 
normality of a saturated solution of the salt. 

Ans. 

140. How many milligrams of Mn+ + can remain dissolved in 100 ml. of a 
solution of pH 8.6 without precipitating Mn(OH) 2 ? 

Ans. 13.5 mg. 

141. From the solubility product of Fe(OH) 8 , calculate the weight of Fe+++ 
in milligrams which must be present in one liter of solution in order to cause 
precipitation of the hydroxide when the hydroxyl-ion concentration is 8.0 X 
10~ 6 mole per liter. 

Ans. 1.2 X 10~ 19 mg. 

142. Given KS.P. MgCO 3 2.6 X lO" 6 ; KS.P. CaCOs = 1.7 X 10~ 8 . In a 
solution 0.20 molar in Ca+ + and 0.20 molar in Mg+ + and with a volume of 
250 ml., which cation would precipitate first bn the slow addition of Na^COa? 
How many milligrams of this cation would still remain in solution when the 
other cation just starts to precipitate? 

Ans. About 1.3 mg. 

143. What are the solubility products of BaF 2 and of BaSO 4 at a certain 
temperature if the solubilities at that temperature are 1.3 grams per liter and 
2.5 X 10~ 3 gram per liter, respectively? A solution has a volume of 100 ml. 
and contains 0.010 mole of Na 2 SO 4 and 0.020 mole of NaF. If BaCl 2 is slowly 
added, which anion will precipitate first? How many milligrams of this ion 
will still remain in solution when the other ion just begins to precipitate? 

Ans. 1.6 X 10-, 1.1 X KT 10 . Sulfate. 0.027 mg. 

144. A saturated solution of magnesium fluoride, MgF 2 , is 1.2 X 10" 3 
molar. What is the solubility product of magnesium fluoride? 

146. What are the solubility products of CaS0 4 and of CaF 2 if the solu- 
bilities are 1.1 mg. per milliliter and 0.016 mg. per milliliter, respectively? 
How many milligrams of calcium ions can remain in 100 ml. of a solution that 
is 0.50 molar in fluoride ions? 

146. If the solubility products of BaC 2 O 4 and of Ba(IO 3 ) 2 are 1.7 X 10~ 7 
and 6.0 X 10~ 10 , respectively, what is the solubility of each salt in milligrams 
per liter? 



66 CALCULATIONS OF ANALYTICAL CHEMISTRY 



147. Mercurous iodide, Hg2l2, dissociates into B^" 1 "* and 21"". Its solu- 
bility product is 1.2 X 10" 28 . How many milligrams of the salt dissolve in 
250 ml.? 

148. The normality of a saturated solution of cerous iodate, Ce(I0 3 )3, is 
5.7 X 10~~ 3 . What is the solubility product of the salt? How many milligrams 
of cerous ions can remain dissolved in 500 ml. of a solution that is 0.30 molar 
in iodate ions? 

149. If A grams of Ba 3 (AsO 4 ) 2 dissolve in 500 ml., express the solubility 
product of Ba 3 (AsO) 4 )2 in terms of A. 

160. If the solubility product of Ag 2 Cr 2 O 7 is 2.7 X 10~ u , how many milli- 
grams of silver will be present in solution when excess salt is shaken with 
250 ml. of water until equilibrium is reached? 

151. If a saturated solution of Pb 3 (PO 4 ) 2 is A normal, what is the solubility 
product of Pb 3 (PO 4 ) 2 ? 

152. If the solubility product of Ag 2 OrO 4 is A, what (in terms of A) is the 
normality of a saturated solution of Ag 2 CrO 4 ? How many grams of chromium 
can remain dissolved (as CrO 4 == ) in 500 ml. of a solution that is B molar in 
silver ions? 

153. Mercurous chloride, Hg 2 Cl 2 , ionizes as follows: Hg 2 CI 2 > Hg 2 ++ + 
2C1~. If its solubility product is 1.1 X 10~ 18 , how many grains of mercurous 
mercury can remain dissolved in 2.00 ml. of a solution that contains one gram- 
milliequivalent weight of chloride ions? 

154. How many grams of FeCl 3 could be present in 200 ml. of an acid solu- 
tion with a pH value of 3.0 without causing a precipitation of Fe(OH) 3 ? 

155. Show by calculation from the solubility product of Ag2SO 4 whether or 
not this compound would be suitable as a final precipitate in the detection or 
determination of silver. What would the concentration of sulfate ions theo- 
retically have to be in solution so that not more than 30 mg. of silver would 
remain unprecipitated in 500 ml. of solution? 

156. What is the ratio of the concentrations of Br~ and Cl~ in a solution in 
which sufficient AgNO 3 has been added to cause precipitation of both halides? 
Solubility products: AgBr = 5.0 X lO" 13 ; AgCl = 1.0 X 10~ 10 . 

157. Calculate the number of milligrams of CaSO 4 converted to CaCO 3 by 
20 ml. of 2.0 N Na 2 CO 3 solution under equilibrium conditions at a temperature 
at which the solubility products are 6.4 X 10~ 5 and 1.6 X 10~ 8 for CaSO 4 and 
CaCO 3 , respectively. 

158. From the appropriate solubility products show which cation would 
precipitate first on the slow addition of K 2 CrO 4 to 500 ml. of a solution 0.10 
molar in Sr++ and 0.10 molar in Ba++. How many milligrams of this cation 
would still remain in solution when the other cation just starts to precipi- 
tate? 

159. What are the solubility products of CaSO 4 and of CaF 2 if the solubili- 
ties are 1.1 grams per liter and 0.016 gram per liter, respectively? A solution 



EQUILIBRIUM CONSTANTS 67 

has a volume of 250 ml. and contains 0.020 mole of Na2SO 4 and 0.030 mole of 
NaP. If CaCl 2 is slowly added, which anion will precipitate first? How many 
milligrams of this ion will still remain in solution when the other ion just begins 
to precipitate? 

36. Application of Buffered Solutions in Analytical Chemistry. 

Buffered solutions are frequently used in both qualitative and 
quantitative analysis to effect certain separations of elements. A 
familiar case is one in which a solution is buffered, usually either 
with NH 4 OII + NH 4 C1 or with HC 2 H 3 O 2 + NH 4 C 2 H 3 O 2 , and the 
pH value thus brought to such a value that the solubility product 
of the hydroxide of an element (or the hydroxides of a group of 
elements) is greatly exceeded but the solubility products of other 
hydroxides are not reached. 

The composition of many insoluble hydroxides is somewhat 
variable, and they are perhaps more properly called " hydrous 
oxides." Their solubility products are not known accurately and 
numerical values obtained from them should therefore be con- 
sidered as showing only relative orders of magnitude. 

EXAMPLE. The solubility product of Mg(OH) 2 at a certain 
temperature is 3.4 X 1Q- 11 ; that of Fe(OH) 3 is 1.1 X 1Q- 36 . At 
that temperature (a) how many graips of Mg++ and of Fe" HHf 
can remain dissolved in 100 ml. of M/10 solution of NH 4 OH 
(ionization constant = 1.75 X 10~ 5 ); (6) how many grams of Mg"^ 
and of Fe^+ can remain dissolved in 100 ml. of M/10 NH 4 OH 
containing a sufficient amount of dissolved NH 4 C1 to make the 
ammonium-ion concentration 2.0 molar? 
SOLUTION : 
(a) NH 4 OH <=> NH 4 + + OH- 



[NH 4 +][OH-] 
[NH 4 OH] 






Let x = concentration of OH~~ = concentration 
Then 

0.10 x = concentration of undissociated NH 4 OH 
(*)(*) 



0.10 - x 



= 1.75 X 10~ 5 



^ |ft = 1.75 X 10~ 5 (since x is small compared to 0.10) 



68 CALCULATIONS OF ANALYTICAL CHEMISTRY 

x = 1.3 X 10~ s mole per liter 

[Mg++][OH-] 2 = 3.4 X 10-" 

[Mg++](1.3 X 10- 3 ) 2 = 3.4 X 10- 11 

[Mg++] = 2.0 X 10- 6 mole per liter 
= 2.0 X 10- 6 X Ko X 24 -3 
= 4.9 X 10- 6 gram per 100 ml. Ans. 

[Fe+++][OH-] 3 = 1.1X10- 36 
[Fe+++](1.3 X 10- 3 ) 3 = 1.1 X 10- 36 



2.8 X 10- 27 gram per 100 ml. Ans. 



* = 8.8 X 10- 7 
[Mg++](8.8 X 10- 7 ) 2 = 3.4 X 10- 11 



= 106 grams per 100 ml. Ans. 

\Fe+++](S.S X 10~ 7 ) 3 = 1.1 X 10- 36 
1-1 X 10-* _1 X558 



(8.8X10~ 7 ) 3 10 
= 9.0 X 10~ 18 gram per 100 ml. Ans. 

37. Control of Acidity in Hydrogen Sulfide Precipitations. The 

separation of certain elements by precipitation from acid solution 
with H 2 S is effectively used in analytical chemistry, particularly 
in qualitative analysis. Probably the most important factor in- 
fluencing the effectiveness of the separation is the sulfide-ion con- 
centration and its control by the regulation of the hydrogen-ion 
concentration. The concentration of the sulfide ion can be regu- 
lated to such a point that the solubility products of certain sul- 
fides are greatly exceeded while the solubility products of other 
sulfides are not reached. The quantitative effect of the presence 
of acid on the ionization of H 2 S and the calculation of the sulfide- 
ion concentration have been illustrated in Examples I and II of 
Sec. 31 and should be reviewed at this time. 



EQUILIBRIUM CONSTANTS 69 

Solubility products of sulfides are not known precisely, and 
hydrolysis effects and rates of precipitation influence the quan- 
titative aspect of the separation of sulfides. Therefore, in the 
following example and problems of a similar nature, the calculated 
values may not agree well with corresponding values determined 
experimentally, but they do show relative orders of magnitude 
and are useful only in this connection. 

EXAMPLE. How many grams of Zn + + and how many grams 
of Cd++ can remain dissolved in 200 ml. of the solution of H 2 S + 
HC1 mentioned in Example II of Sec. 31 (solubility product of 
ZnS = 1.2 X 10~ 23 ; solubility product of CdS = 3.6 X 10~ 29 )? 
SOLUTION: 

[S-] = 1.5 X 10~ 23 (as calculated) 
[Zn++][S=] = 1.2 X 10- 23 



1 9 
[Zn++] = zr 3 = 0.80 mole per liter 



= 0.80 X 65 X H = 10 grams per 200 ml. Ans. 

[Cd++][S-] = 3.6 X 10~ 29 

= 3.6 X 
J 



1.5X10-* 
= 2.4 X 10~ 6 molc x pcr liter 
= 2.4X 10- 6 X 112 X% 
= 5.4 X 10~ 5 gram per 200 ml. Ans. 

38. Separations by Means of Complex-ion Formation. Certain 
separations in analytical chemistry are effected by making use of 
the equilibrium that exists between a complex ion and its constitu- 
ents. The following cases illustrate the two general ways in which 
this is applied. 

1. When an ammoniacal solution of silver nitrate containing a 
carefully controlled excess of ammonia is added to a mixture of 
iodide and chloride, only silver iodide is precipitated, since most 
of the silver in the solution is as the ammino complex, Ag(NH 3 ) 2 " h , 
and the concentration of Ag+ is too small to exceed the solubility 
product of AgCl but is great enough to exceed the solubility 
product of the more insoluble Agl. 

2. When potassium cyanide is added to an ammoniacal solution 
of copper and cadmium salts, the two ions Cu(CN)3 =s and Cd(CN) 4 *" 
are formed. When hydrogen sulfide is passed into the solution, 



70 CALCULATIONS OF ANALYTICAL CHEMISTRY 

only cadmium sulfide is precipitated, since the degree of dissocia- 
tion of the copper complex is much less than that of the cadmium 
complex. A sufficiently high concentration of Cd ++ is present to 
exceed the solubility product of CdS, but the concentration of Cu+ 
is too low to exceed the solubility product of Cu2S. 

EXAMPLE I. How many grams of silver bromide will dissolve 
in one liter of NH 4 OH if the resulting solution is 2.0 molar in NH 3 ? 
SOLUTION : 

[Ag+][Br~] = 5.0 X 10~ 13 (see Appendix) 



Let x = moles of AgBr dissolved = [Br~] = [Ag(NH 3 ) 2 + ] 

5.0 X 10~ 13 _ , 
x 

(5.0XlO-"/s)(2.0) _ 

u.o A 1U 
X 

Solving, 

x = 5.4 X 10~ 3 molar 
5.4 X 10~ 3 X AgBr = 1.0 grain. Ans. 

EXAMPLE II. A solution 0.10 molar in Cu ++ and 0.10 molar 
in Cd++ is treated with NH 4 OH and KCN, forming Cu(CN) 3 === 
and Cd(CN) 4 ^. The solution is 0.50 molar in excess CN~ ions. 
If H 2 S is passed into the solution to give a sulfide-ion concentra- 
tion of 0.010 molar, will Cu 2 S or CdS precipitate? 
SOLUTION : 



n J = 5 - x 10 ~ 28 ( see Appendix) 

NJ3 J 



tCu + l(0.60) _ 5 Q 1Q _ 28 
0.10 ~ 5 ' X 1U 

[Cu+] = 4.0 X 10- 28 
Therefore, 

[Cu+] 2 [S=] = (4.0 X 1(^ 
= 1.6X10- 57 

The solubility product of Cu 2 S (= 1.0 X 10~ 50 ) is greater than this 
value. Hence CuzS will not precipitate. Ans. 



EQUILIBRIUM CONSTANTS 71 

[Cd++][CN-]* _ 

~ * 



[Cd++](0.50)* _ 

0.10 ~ X 

[Cd++] = 2.2 X 10~ 17 
Therefore 

[Cd++][S~] = (2.2 X 10- 17 )(0.01) 
= 2.2 X 10- 19 

The solubility product of CdS (= 3.6 X 10~ 29 ) is less than this 
value. Hence CdS will precipitate. Ans. 

Problems 

(See Appendix for solubility products and ionization constants.) 

160. How many grams of Mg++ could remain dissolved [i.e., unprecipitated 
as Mg(OH) 2 ] in a liter of 0.2 M NH 4 OH, and how many grams of Mg+ + could 
remain dissolved in a liter of 0.2 M NH 4 OH containing enough dissolved 
NH 4 C1 to make the ammonium-ion concentration 1 molar? [Ks.p. Mg(OH)-> = 
3.4 X 10- 11 .] 

Ans. 2.4 X 10~ 4 gram, 68 grams. 

161. How many milligrams of Fe ++ + could remain dissolved [i.e., unpre- 
cipitated as Fe(OII)3] in 100 ml. of a solution 2.0 normal in acetic acid and 
containing a sufficient amount of sodium acetate to make the acetate-iori con- 
centration 0.15 molar? IK S .P. Fe(OH) 3 = 1.1 X 10~ 36 .] 

Ans. 0.095 nig. 

162. Calculate from appropriate equilibrium constants the number of grams 
of zinc and of cadmium that can remain dissolved in 1,500 ml. of a solution 
that contains 0.05 mole of dissolved H 2 S and is 0.30 N in hydrogen ions. 

Ans. 29.0 grams, 1.5 X 10" 4 gram. 

163. By saturating with hydrogen sulfide 350 ml. of a solution that is 0.010 
molar in a certain trivalent element and 1.0 molar in hydrogen ions, all but 12 
millimoles of the element precipitates as sulfide. What is the approximate 
solubility product of the sulfide of the element? (Solubility of H 2 S = 0.10 
molar.) 

Ans. 1.5 X 10~ 72 . 

164. How many formula weights of chloride ion must be introduced into 
a liter of a tenth-molar solution of NaAg(CN) 2 in order for AgCl to start to 
precipitate? [Dissociation constant of Ag(CN) 2 ~~ = 1.0 X 10~ 21 ; solubility 
product of AgCl = 1.0 X lO' 10 .] 

Ans. 0.0033 F.W. 

166. How many formula weights of silver iodide will dissolve in one liter 
of NH 4 OII which is 6.0 molar in NH 3 ? [Dissociation constant of Ag(NH 3 ) 2 + = 



72 CALCULATIONS OF ANALYTICAL CHEMISTRY 

6.8 X 10~ 8 ; solubility product of Agl = 1.0 X 10~ 16 .] (Hint: In the resulting 
solution [Ag(NH 8 ) 2 +} = [I"].) 
Ans. 2.4 X 10~ 4 F.W. 

166. A solution 0.080 molar in AgNO 3 is treated with Na 2 S2O 8 which con- 
verts practically all of the Ag+ into Ag(S 2 Os)2 s . If the solution contains 
sufficient excess thiosulfate to make the S 2 O 3 ~ 0.20 molar, how many grams 
of I~ per liter could be present without causing a precipitation of Agl? 

Ans. 0.16 gram. 

167. What is the maximum molar concentration of sulfide ion in a solution 
0.20 molar in Cd(NH 3 ) 4 Cl2 and 2.0 molar in NH 8 without forming a precipitate 
of cadmium sulfide? 

Ans. 1.2 X 10- 20 . 

168. What must be the maximum pll value of a solution in order that 
0.500 gram of Mg++ in 100 ml. will remain unprecipitated as Mg(OH) 2 ? 
[Solubility product Mg(OH) 2 = 3.4 X lO" 11 .] How many grams of Fe ++ + 
could remain dissolved in such a solution? 

169. How many milligrams of Mii ++ could remain unprecipitated as 
Mn(OH) 2 in 500 ml. of 0.10 M NH 4 OH, and how many milligrams of Mn + + 
could remain dissolved in 500 ml. of 0.10 M NH 4 OII containing sufficient 
dissolved NH 4 C1 to make the ammonium-ion concentration 2.0 molar? 

170. How many milligrams of Fe+ ++ could remain unprecipitated as 
Fe(OH) 8 in 250 ml. of a solution 1.5 molar in acetic acid buffered by that 
amount of dissolved sodium acetate that makes the acetate-ion concentration 
0.20 molar? 

171. If the solubility product of Bi2S 8 is 1.6 X 10~ 72 , find the weight of 
bismuth ions that must be present in a liter of solution to cause precipitation 
of Bi2S 8 in a solution which is 0.10 molar in 1X28 and contains 0.010 mole of II 4 
per liter. 

172. What is the maximum pH value that 100 ml. of a solution containing 
0.0050 gram of PbCl2 can have so that on saturating the solution with II 2 S no 
lead sulfide will precipitate? (Saturated solution of H 2 S = 0.10 molar.) 

173. A solution of 1.2 grams of ZnSO 4 .7H 2 O in 500 ml. of dilute acid is 
saturated with H 2 S. The resulting solution is found to be 0.10 molar in H 2 S 
and 0.050 molar in H+ ions. What fraction of the zinc has been precipitated 
as ZnS? What maximum pll value should the solution have in order for no 
precipitate to form if the concentrations of zinc salt and H2S are as above? 

174. If 50 ml. of 0.010 molar AgN0 3 and 50 ml. of 3.0 molar NH 4 OH are 
mixed, what is the resulting concentration of Ag+ ions? How many moles of 
Cl~ would have to be introduced before precipitation of AgCl would take 
place? [lonization constant of Ag(NH 3 ) 2 + = 6.8 X 10"" 8 ; solubility product 
of AgCl = 1.0 X lO" 10 .] 

176. If a solution is 0.050 molar in K 2 HgI 4 and 1.5 molar in I", show by 
calculation whether or not a precipitation of HgS would be expected if the 
solution is made 1.0 X 10"" 15 molar in sulfide ions. 



EQUILIBRIUM CONSTANTS 73 

176. How many grams of S"" can be present in a liter of a solution containing 
0.10 F.W. of Cd(NH 3 ) 4 Cl 2 and 1.5 mole of NH 8 without forming a precipitate 
of CdS? 

177. A solution containing 0.10 F.W. of MgCl 2 and 0.20 F.W. of CaCl 2 
would require a total of how many milliliters of N/2 H 2 C 2 O 2 solution in order 
to form the complex ion Mg(C 2 C>4) 2 "" and precipitate the calcium as CaC 2 O 4 . 
H 2 0? 

178. How many grams of silver bromide will dissolve in one liter of NKUOH 
which is 1.5 molar in NH 3 ? [Dissociation constant of Ag(NH 3 ) 2 + = 6.8 X 
10~ 8 ; solubility product of AgBr = 5.0 X 10~ 13 .] (Hint: In the resulting 
solution [Ag(NII 3 ) 2 + ] = [Br~].) 

39. Distribution Ratio. Occasionally in analytical chemistry 
the greater part of a solute is removed from aqueous solution by 
shaking the solution with an organic solvent in which the solute 
is much more soluble. In qualitative analysis the removal of 
liberated bromine and iodine by means of carbon tetrachloride 
in the test for bromide and iodide occurs in many schemes of 
analysis. In quantitative analysis ferric chloride is often extracted 
in greater part from hydrochloric acid solution by means of ethyl 
ether or isopropyl ether. This is of value in the analysis of certain 
constituents in iron alloys where a high concentration of ferric 
ions in the solution is undesirable. Certain hydrolytic separations 
(e.g., titanium from iron) are more readily carried out if a prelim- 
inary extraction of the iron is made. 

The distribution law states that when a solute is in simultaneous 
equilibrium with two mutually insoluble solvents the ratio of the 
concentrations of the solute in the two solvents is a constant at 
a given temperature regardless of the volumes of solvents used 
or of the quantity of solute present: 

Concentration of x in solvent A __ 
Concentration of x in solvent B 

The ratio K is called a distribution ratio and is determined 
experimentally for each solute and each pair of mutually insoluble 
liquids. The value is a true constant only in the case of perfect 
solutions and perfectly immiscible solvents, but it is very nearly 
a constant at a given temperature for the dilute solutions ordi- 
narily encountered in analytical chemistry. The law applies only 
to a particular species of molecule; that is, the solute must be in 
the same condition in the two phases. The law does not hold, 



74 CALCULATIONS OF ANALYTICAL CHEMISTRY 

for example, if the solute is ionized in one solvent and not ionized 
in the other, or if it is as associated molecules in one solvent and 
not in the other. 

EXAMPLE. An aqueous potassium iodide solution has a volume 
of 100 ml. 'and contains 0.120 gram of dissolved iodine. Assuming 
that the distribution ratio of iodine between carbon tetrachloride 
and an aqueous solution of potassium iodide is 85 at 25C., how 
many grams of iodine will remain in the aqueous phase if the 
above solution is shaken with 25 ml. of carbon tetrachloride? 
SOLUTION: 

Let x = grams iodine remaining in II 2 O phase 

Then 

0.120 x = grams iodine in CCL* phase 

120 x 

Concentration of iodine in CC1 4 phase = ~ ^= - grams per ml. 

.Zo 

X 

Concentration in KUO phase = y^: grams per ml. 



(0.120 -s)/25 



Solving, 

x = 0.00539 gram. Ans. 

Problems 

173. If 0.568 gram of iodine is dissolved in 50 ml. of carbon tetrachloride 
and the solution is shaken at a certain temperature with 500 ml. of water, it is 
found that the aqueous layer contains 0.0592 gram of iodine. Calculate the 
distribution ratio of iodine at that temperature between the two solvents, in 
both of which it exists as I 2 molecules. 

Ans. 86.0 

180. At 20C. the distribution ratio of a certain organic acid between water 
and ether is 0.400. A solution of 5.00 grams of the acid in 100 ml. of water is 
shaken successively with three 20-ml. portions of water-saturated ether. Cal- 
culate the number of grams of acid left in the water. Also calculate the number 
of grams of acid that would have been left in the water if the solution had been 
shaken with a single 60-ml. portion of ether. 

Ans. 1.48 grams. 2.00 grains. 

181. If 90 ml. of an aqueous solution containing 1.00 millimole of bromine 
are shaken at 25C. with 30 ml. of a certain organic solvent, 0.128 gram of the 
bromine is extracted from the aqueous layer. What is the distribution ratio? 
What percentage of the bromine would have been extracted by two successive 



EQUILIBRIUM CONSTANTS 75 

extractions with 15-ml. portions of the solvent? Assume bromine to be as 
diatomic molecules in both solvents. 

Ans. 12.0. 88.89 per cent. 



182. The distribution ratio of bromine between carbon tetrachloride and 
water is 29.0 at 25C. If a certain aqueous solution of bromine is shaken with 
one-half its volume of carbon tetrachloride, what percentage of the bromine 
is removed from the aqueous phase? 

183. If a 0.0010 molar solution of bromine in solvent A is shaken with one- 
tenth of its volume of solvent B, 78.0 per cent of the bromine remains in A. If 
at the same temperature a solution of 0.0300 gram of bromine in 50 ml. of 
solvent B is shaken with 20 ml. of solvent A } how many grains of bromine 
remain in B? Assume A and B to be immiscible and bromine to be as diatomic 
molecules in both solvents. 

184. What explanation can you give to account for the fact that the dis- 
tribution ratio at 25C. of acetic acid between benzene and water is not even 
approximately constant, even at low concentrations of acetic acid? 

185. In certain methods of analysis iron is removed from hydrochloric acid 
solution by repeated extraction with either ordinary ether or with isopropyl 
ether. If 36 ml. of an aqueous solution of FeCl 3 + HC1 are shaken with 18 ml. 
of ether (previously saturated with IIC1), 94 per cent of the iron is removed 
from the aqueous layer. What is the distribution ratio of iron between the 
two solvents and what per cent of the iron would have been removed if the 
initial solution had been extracted with two separate 9-ml. portions of ether? 



CHAPTER VI 
OXIDATION POTENTIALS 

40. Relation of the Electric Current to Oxidation-reduction 
("Redox") Reactions. Experiment shows that at ordinary con- 
centrations free chlorine or bromine will oxidize ferrous ions 
(2Fe++ + C1 2 -> 2Fe+++ + 2C1~), but free iodine will not. Con- 
versely, iodide ions will reduce ferric ions (2Fe +++ + 2I~ * 
2Fe + + + 12) but chloride or bromide will not. Hydrogen ions at 
ordinary concentrations will oxidize metallic zinc (Zn + 2H+ -> 
Zn* 4 " + H2) but will not oxidize metallic copper. 

To be able to predict whether or not a given pair of oxidizing 
and reducing agents will or will not mutually interact to an ap- 
preciable extent is of considerable importance, particularly in 
qualitative analysis, and tables showing relative tendencies for 
substances to be oxidized or reduced, if used properly, are of great 
value. 

In the light of the modern concept of the structure of atoms, 
oxidation and reduction may be defined in terms of transfer of 
electrons. An element is oxidized when it loses electrons; an ele- 
ment is reduced when it gains electrons. 

Redox reactions can be brought about by the application of an 
electric current; conversely, an electric current can be obtained 
from oxidation-reduction processes. The electrolysis of a solution 
of sodium chloride is an example of the first class. At the anode, 
negative chloride ions are oxidized to free chlorine gas; at the 
cathode, positive hydrogen ions from the water are reduced to free 
hydrogen gas. The voltaic cell is an example of the second class. 

41. Specific Electrode Potentials. Suppose we have on the end 
of a platinum wire a platinum foil covered with platinum black. 
Suppose further that the foil is immersed in a solution of acid 
that is one molar in hydrogen ions and that pure hydrogen gas 
at one atmosphere pressure continually bubbles over the foil. 
Such a setup is called a normal hydrogen electrode. It may be 
represented graphically, thus: H2(l atmosphere), 2H + (1 molar) I 

76 



OXIDATION POTENTIALS 77 

Pt. The platinum is chemically inert, but an equilibrium exists 
between the hydrogen gas and the hydrogen ions, thus: H 2 <=* 
2H+ + 2, the symbol e representing an electron. Suppose now 
we have a strip of metallic zinc immersed in a solution of zinc 
sulfate that is one molar in zinc ions. Equilibrium exists between 
the metal and its ions, thus: Zn <= Zn++ + 2e. If the two elec- 
trodes are connected by means of a wire and the two solutions 
are connected by means of a capillary tube containing an elec- 
trolyte, a current of 0.76 volt will flow through the wire and solu- 
tion. The "plus to minus " direction of the current in the wire 
is from the hydrogen electrode to the zinc electrode and in the 
solution from the zinc electrode to the hydrogen electrode. The 
flow of electrons in the wire is from the zinc electrode to the hy- 
drogen electrode, and in the solution it is from the hydrogen 
electrode to the zinc electrode. At the same time, metallic zinc 
is oxidized and hydrogen ions are reduced, the net reaction being 
represented by the equation Zn + 2H+ > Zn 4 " 1 " + H 2 . We there- 
fore have a voltaic cell made up of two half cells, and the entire 
system may be represented thus: 

Zn Zn++(l molar) 11 H+(l molar), H 2 (l atmosphere) Pt 



In representing cells in this way, a single line represents a junc- 
tion between an electrode and a solution. A double line denotes 
a junction between two solutions, and it is assumed that the 
small potential difference between the solutions has been corrected 
for in formulating the total e.m.f. of the cell. 

It should also be noted that oxidation always takes place at 
the anode; reduction always takes place at the cathode. The pas- 
sage of electrons through the wire is from anode to cathode; elec- 
trons pass through the solution from cathode to anode. 

In similar fashion, a copper electrode dipping in a solution of 
copper sulfate that is one molar in copper ions can be connected 
to a normal hydrogen electrode. A current of 0.34 volt will flow 
through the wire. The passage of electrons in the wire is from 
the hydrogen electrode to the copper electrode. 

If now we connect the above copper half cell with the above 
zinc half cell, we obtain a voltaic cell that is represented thus: 

Cu I Cu++(l molar) II Zn++(l molar) I Zn 



78 CALCULATIONS OF ANALYTICAL CHEMISTRY 

It will be found that a current of 1.10 volts will be generated. 
The passage of electrons in the wire is from the zinc to the copper 
and in the solution from the copper to the zinc. At the same time 
metallic zinc is oxidized to zinc ions, and the copper ions are re- 
duced to metallic copper, the net reaction being 

Zn + Cu++ -> Cu 



It is difficult to determine absolute potential differences be- 
tween electrodes and solutions; but, since we are usually con- 
cerned only with differences of potential, we can refer electrode 
potentials to some common standard. The normal hydrogen elec- 
trode is arbitrarily given the value of zero, and other electrode 
potentials are referred to it. The molar electrode potential or 
specific electrode potential of zinc (i.e., the relative potential be- 
tween metallic zinc and a one-molar solution of zinc ions) is 
0.76 volt; the specific electrode potential of copper is + 0.34 
volt. Giving the zinc potential the negative sign and the copper 
potential the positive sign is again purely arbitrary. It is perhaps 
more from the point of view of the physicist who is primarly inter- 
ested in the outer circuit of a cell, than from the point of view of 
the chemist who is concerned with the changes within the solu- 
tion. To the physicist, copper is positive to zinc since in the 
above cell the positive to negative direction of the current in the 
wire is from the copper to the zinc. Chemists are also adopting 
this sign convention although many texts and reference books use 
the opposite convention. 

In this book, specific potentials will be denoted by the symbol 
E, and a table of such potentials is given in the Appendix. When 
applied to an active metallic electrode, the numerical value refers 
to the potential at 25C. between a metal and a one-molar solu- 
tion of its ions relative to the potential between hydrogen gas at 
one atmosphere pressure and a one-molar solution of hydrogen 
ions. 

The e.m.f. of a cell is the difference between the potentials of 
its two half cells, or E = EI E 2 . In the case of the above- 
mentioned cell, E = E l - E* = Ecu* - E ZU Q = +0.34 - (-0.76) = 
1.10 volts. 

Electrode potentials are not limited to those between elements 
and their ions. They also apply to potentials between ions at 



OXIDATION POTENTIALS 



79 



two states of oxidation. Thus, as shown in the potential table 
in the Appendix, the specific potential between ferrous and ferric 
ions (Fe++ =* Fe"*" 4 " 4 " + e) is +0.748 volt, indicating that a current 
of that voltage would flow through the following cell: 



Pt 



M), 



M) 



M), H 2 (l atm.) Pt 



Electrons would pass through the solution from the ferrous-ferric 
half cell to the hydrogen electrode (i.e. y from left to right as written 
above). They would enter the solution from the wire at the fer- 
rous-ferric half cell. Ferric ions would be reduced to ferrous ions; 
hydrogen gas would be oxidized to hydrogen ions. 

Similarly, the specific potential between chromic ions and di- 
chromate ions in the presence of acid (20+++ + 7H2O =* Cr^O^ + 
14H+ + 6e) is +1.30 volt which is the voltage of the following 
cell: 

Cr+++(l M) 

" H+(l M), H 2 (l atm.) 



Pt 



M) 



H+(l M) 



Pt 



The e.m.f. of the following cell: 



Pt 



Fc++(l M) 
Fe+++(l M) 


Cr+++(l M) 
Cr 2 O7 = (l M) 
H + (l M) 



Pt 



would be the algebraic difference between the two half cells com- 
prising it, or +0.748 (+1.30) = 0.55 volt. The negative sign 
indicates that the passage of electrons through the solution is from 
right to left as written above, and through the wire from left to 
right. The ferrous ions are therefore oxidized and the dichromate 
ions reduced during the process. The over-all reaction would be 



Cr 2 C>7- + 14H+ 



7H 2 



42. Rules for Writing Equations for Half-cell Reactions. In 

writing and balancing equations for half-cell reactions the follow- 
ing steps should be followed: 

1. Write the reduced form of the element that changes its oxida- 
tion number on the left-hand side of the equation; write the 
oxidized form on the right-hand side. If necessary, balance the 
number of atoms of the element by inserting the proper coefficients. 



80 CALCULATIONS OF ANALYTICAL CHEMISTRY 

2. On the right-hand side of the equation introduce that number 
of electrons equal to the total change in oxidation number of the 
element. 

3. If necessary, introduce sufficient hydrogen ions (if the reac- 
tion takes place in acid solution) or hydroxyl ions (if the reaction 
takes place in basic solution) to balance the electrical charges. 
Remember that each electron symbol represents a negative charge. 

4. If necessary, introduce water molecules into the equation to 
balance the hydrogen and oxygen atoms. 

EXAMPLE. Write balanced half-cell reactions for the following 
changes (a) VO++->V0 3 ~ (acid solution); (6) Cr+++ - O 2 7 = 
(acid solution); (c) Mn++ MnO 2 (basic solution). 

Following the above four steps, the results in each case are as 
follows: 

(a) 1. VO++->VOr 

2. VO++ - V0 3 - + e (change = 5-4=1) 

3. VO++ -> V0 3 - + 4H+ + e 

4. VO-*+ + 2H 2 -> V0 3 - + 4H+ + e 



(6) 1. 2Cr+++--Cr 2 7 ~ 

2. 2O+++ -> Cr 2 O 7 - + 6e [change = (6 - 3) X 2 = 6] 

3. 2O+++ -> Cr 2 Or + 14H+ + 6e 

4. 2Cr+++ + 7H 2 O - Cr 2 O 7 = + 14H+ + 6e 



(c) 1. Mn++ -+ MnO 2 

2. Mn++ -> MnQ 2 + 2e 



3. Mn++ + 40H- -> Mn0 2 + 2e 



4. Mn-^ + 40H- -> MnO 2 + 2H 2 + 2e 

43. Oxidation-reduction Equations in Terms of Half-cell Reac- 
tions. In order to write an ordinary redox equation in terms of 
half -cell reactions the appropriate couples are merely written one 
below the other and subtracted in algebraic fashion. Since electron 
symbols should not appear in the net equation, it is frequently 
necessary to multiply one or both half-cell equations by a factor 
in order that the electrons may " cancel" out. This is illustrated 
in the following examples. Obviously the potential of the half- 
cell reaction is not affected by such multiplication. 

The oxidation of ferrous ions by chlorine can be written: 



OXIDATION POTENTIALS 81 

Fe++ - FC+++ + e (Ef = +0.747) 

or 

(1) 2Fe++ = 2Fe^ + 2e 

(2) 2C1- = C1 2 + 2e (# 2 = +1.359) 
(l)-(2) 2Fe++ + C1 2 = 2Fe+++ + 2C1~ 

The oxidation of ferrous ions by dichromate in the presence of 
acid can be written: 

(1) 6Fe++ = 6Fe+++ + 6e (EJ = +0.747) 

(2) 2Cr+++ + 7H 2 O = Cr 2 7 = + 14H+ + 6e (E 2 = +1.30) 

(l)-(2) 6Fe++ + Cr 2 O 7 - + 14H+ = 6Fe+++ + 2Cr+++ + 7II 2 O 

The oxidation of stannous ions in the presence of acid by per- 
manganate can be written: 

(1) 5Sn++ = 5Sn++++ + We (Ei* = +0.13) 

(2) 2Mn++ + 8H 2 = 2MnOr + 16H+ + lOe (E 2 = +1.52) 



2MnO 4 - + 16H+ = 5Sn +++ + + 2Mn++ + 8H 2 O 

If concentrations are all 1 molar, the net potentials of the above 
three illustrations are the algebraic differences between the specific 
potentials corresponding to the two half^cell reactions, namely, 

E = EJ- E z = (+0.747) - (+1.359) = -0.612 volt 
E = Ef - E 2 = (+0.747) - (+1.30) = -0.55 volt 
E^ES- E 2 = (+0.13) - (+1.52) = -1.39 volt 

In cases like the above, if the algebraic difference between the elec- 
trode potentials, as written, is negative, the net reaction can be ex- 
pected to go as written (i.e., from left to right). If the algebraic dif- 
ference is positive, the reaction will not go as written but can be 
expected to go from right to left. Thus, 

(1) 2Fe++ = 2Fe+++ + 2e (+0.747) 

(2) 21- = I 2 + 2e (+0.535) 

(l)-(2) 2Fe-H- + I, = 2Fe+++ + 2l~ 

[(+0.747) - (+0.535) = +0.212 volt] 

This reaction at 1 molar concentrations will not take place from 
left to right as written, but goes in the opposite direction (2F 

21- - 2Fe++ + I 2 ). 



82 CALCULATIONS OF ANALYTICAL CHEMISTRY 

The behavior of metals with acids can be treated in the same 
way. Metallic zinc dissolves in 1 M HCL 

(1) Zn = Zn++ + 2e (-0.758) 

(2) H 2 = 211+ + 2 (0.00) 

(l)-(2) Zn + 2H+ -* Zn++ + H 2 (-0.758) 
Metallic copper does not dissolve in HC1. 

(1) Cu = Cu++ + 2e (+0.344) 

(2) H 2 = 2H+ + 26 (0.00) _ 

(l)-(2) Cu + 2H+ - Cu++ + H 2 (+0.344) 

Both zinc and copper dissolve in 1 M HNOa. In this case two 
oxidizing agents are present, namely H+ and N0a~, but the nitrate 
ion has the greater oxidation potential and is the oxidizing agent 
in both cases. 



(1) 3Cu = 3Cu^ + 6e (+0.344) 

(2) 2NO + 4H 2 O - 2NO 3 - + 8H+ + 6e (+0.94) 

(l)-(2) 3Cu + 2N0 3 ~ + 8H+ - 3Cu++ + 4H 2 O 

[(0.344) - (0.94) = -0.60 volt] 

The potential table therefore shows relative tendencies for sub- 
stances to lose or gain electrons. Substances at the top left of 
the table lose their electrons most readily and gain them least 
readily. They are therefore the strongest reducing agents. Simi- 
larly, the oxidizing agents at the bottom of the table are the 
strongest; those at the top are the weakest. 

Such predictions as given above must be applied cautiously. 
In a few cases reactions that should proceed according to the rela- 
tive positions in the potential series do so at such a slow rate that 
they are almost negligible. More important still, as shown in 
the next section, the concentration of each component of an oxida- 
tion-reduction equilibrium affects the value of the potential. A 
substance may be present that is capable of forming a complex 
ion with one of the components of the half-cell equilibrium and 
thus reduces the concentration of that component to a point where 
it no longer reacts. Thus, the potential of the equilibrium Sn++ <= 
g n ++++ 4. 2e is greatly affected by the presence of chloride ions 
which form SnCle" ions with the stannic tin. In a few cases pre- 



OXIDATION POTENTIALS 83 

cipitation effects interfere in the same way. For example, ac- 
cording to the table, iodide ions should reduce silver ions to metallic 
silver (2Ag+ + 2I~ > 2Ag + I 2 ). Actually, a precipitation of silver 
iodide takes place instead (Ag+ + I~ > Agl) and the concentra- 
tion of Ag+ in the residual solution is made too small to be affected 
by excess iodide. In the case of a few metals, passivity effects 
may occur. Pure aluminum should dissolve readily in nitric acid 
(Al + NO 3 ~ + 4H+ -> A1+++ + NO + 2H 2 O). Actually it does not 
do so, probably because of the formation of a protective coating 
of oxide on the surface of the metal. 

44. Relation between Electrode Potential and Concentration. 
When the prevailing concentrations are not 1 molar, the electrode 
potentials are no longer molar electrode potentials but can be cal- 
culated from them. From considerations of free energy it can be 
shown that at 25C. electrode potentials can be calculated from 
the following formula: 

, 0.0591 , ,_ . 

E = E Q H log M l 

n 

where E Q = molar electrode potential 

n = number of faradays involved in the change 
log = common logarithm 

M = ratio obtained by dividing the prevailing molar con- 
centrations of the oxidation products of the reaction 
by the prevailing molar concentrations of the reacting 
substances, each concentration being raised to a power 
equal to the coefficient of the substance in the equation 
representing the reaction taking place in the half cell. 
In expressing M, reactions should be written as oxida- 
tions, and, as in the case of mass-action expressions, 
concentrations of water and of solid substances ar 
omitted. Gases are expressed in terms of partial pres- 
sures (in atmospheres) 

1 As stated in Sec. 41, in some tables of electrode potentials the signs are 
opposite to those given in this book; that is, the numerical values of those 
potentials listed above hydrogen are plus, and those below hydrogen are minus. 
If this alternative system is used, the given formula becomes E = E 

log M. The consistent use of either system leads to correct results. 



84 



CALCULATIONS OF ANALYTICAL CHEMISTRY 



In calculations of electrode potentials, as in the case of calcula- 
tions of equilibrium constants, activities rather than concentra- 
tions should be used for precise results (see Sec. 32). Values for 
specific electrode potentials should therefore be for unit activity 
rather than for molar concentration. Analytical calculations in 
this particular field, however, do not require a precision greater 
than one or two significant figures, and the use of activities and 
activity coefficients (which are not always easily determined) can 
be dispensed with. 

EXAMPLE I. Find the e.m.f. at 25O. of the cell 



Pt 

SOLUTION : 

(1) C 



Cc +++ (0.10 molar) 
Ce ++++ (0.00010 molar) 



Fe++(0.010 molar) 
Fe+++(1.0 molar) 



Pt 



(2) Fe++ = Fe+++ + 



Subtracting (2) from (1) 



= 1.45 + 0.0591 log 10-' 
= 1.27 





Fe++ 



= 0.747 + 0.0591 log 100 
= 0.805 



= E L -Es= 1.27-0.8G5 
= 0.40 volt. Ans. 



The positive sign shows that the passage of electrons in the 
solution is from left to right. The ferrous ions are, therefore, 
oxidized, and the eerie ions are reduced. The reaction Ce 4 " 4 " 1 " + 
Fe+ ++ = Ce 4 ^"^ 4 " + FG ++ proceeds from right to left. The eerie ions 
are reduced, and the ferrous ions are oxidized until the concentra- 
tions so adjust themselves that no current flows. 

In this example the concentrations given are those of the simple 
cations. These cations may often be in equilibrium with complex 
ions, such as FeOLr, Ce(SO 4 ) 3 ^, etc. 

EXAMPLE IT. What is the e.m.f. of the half cell represented 
by the following equilibrium: 2Cr+++(0.20 M) + 7II 2 O <= Cr 2 7 - 
(0.30 M) + 14H+(2.0 M) + Oc? 



OXIDATION POTENTIALS 85 

SOLUTION: 



0.0591 , (0.30) (2.0) 14 



= 1.46 volts. Ans. 
EXAMPLE TIL What is the e.m.f. of the cell 

Cu Cu f +(0.010 molar) || Zn++(1.0 X 10~ 6 molar) Zn? 
SOLUTION: 

El = Ecu" + - log [Cu++] 



= +0.34 + log (0.010) 

Zj 

= +0.28 volt 



= -0.76 + - log^l.O X 10- 6 ) 

Zi 

= -0.94 
E = h\-E 2 = 1.22 volts. ^In^. 

P]XAMPLB IV. Calculate the e.m.f. of the following cdl: 
Cu Cu ++ (2.0 M) | Cu++(0.010 M) Cu 



This type of cell, made up of the same half-cell equilibrium (Cu ^ 
C -u 4 "* + 2e, in this case) but with the ions at two different con- 
centrations, is known as a concentration cell. 
SOLUTION: 



E, = 0.344 -f ~ log 2.0 

z 

o nwi 
^ 2 = 0.344 + ^P log 0.010 



~ (log 2.0 - log 0.010) 
= 0.068 volt. Ans. 



86 CALCULATIONS OF ANALYTICAL CHEMISTRY 

45. Calculation of the Extent to Which an Oxidation-reduction 
Reaction Takes Place. All reversible reactions proceed in one 
direction or the other until equilibrium conditions are reached, at 
which point the two rates of reaction are equal. 

During the progress of an oxidation-reduction reaction in a cell, 
the concentrations of the reacting substances are steadily decreas- 
ing and those of the products are increasing. The voltage of the 
cell decreases steadily until equilibrium is reached, at which point 
no current flows. The potentials of the two half cells making up 
the cell are therefore equal at this point of equilibrium. In order 
to calculate the extent to which an oxidation-reduction reaction 
takes place, it is only necessary to express the reaction as two half- 
cell reactions and to express an equality between the two electrode 
potentials. 

EXAMPLE I. When excess metallic aluminum is added to a 
solution 0.30 M in cupric ions, what is the theoretical concentra- 
tion of CU++ after equilibrium is reached (2A1 + 3Cu++ + 2A1+++ + 
3Cu)? 

SOLUTION: Experiment shows that the reaction is practically 
complete. 

[A1+++] = 0.20 M (3 moles of CU++ give 2 moles of A1+++) 
[Cu++] = x M 

. - n , 0.0591 , n on IA9>I>II 0.0591 , 

-1.70 H 5 log 0.20 = +0.344 + = log x 

o Zi 

log x = -69 

x = 1 X 10- 69 mole. Ans. 

EXAMPLE II. A solution is prepared so as to be initially 0.000 
molar in Fe + +, 0.10 molar in Cr 2 7 = and 2.0 molar in II + . After 
the reaction, what would be the approximate concentration of the 
Fe++ remaining (6 Fe++ + Cr 2 O 7 = + 14H+ <= 6F6+++ + 2Cr-H+ + 
7H 2 0)? 

SOLUTION: Experiment shows that the reaction is practically 
complete. Since, according to the equation, 0.060 mole of Fe++ 
would react with 0.010 mole of Cr 2 O 7 = and 0.14 mole of H+, the 
latter two are initially present in excess and the concentration of 
Fe ++ is the limiting factor. At equilibrium, ^6+++] = 0.060; 
[Fe 4 -*-] = x; [Cr+++] = 0.020; [Cr 2 7 ~] = 0.10 - 0.010 = 0.090; 
[H+] = 2.0 - 0.14 = 1.86. The two half-cell equilibria are 



OXIDATION POTENTIALS 87 

6Fe++(z M) *= 6Fe+++(0.060 M) + 6e 

2Cr+++(0.020 M) + 7H 2 O * 

Cr 2 O 7 =(0.090 M) + 14H+(1.86 M) + 6 



+0.747 + log . +1.30 + log 



, A _._, 0.0591 , 0.060 10A . 0.0591, (0.090) (1. 86) " 
+0.747 + -p- log = 1.30 + -g- log ; 

Solving, 

a: = 1 X 10~ 17 . Ans. 

46. Calculation of Equilibrium Constant from Electrode Po- 
tentials. A mass-action constant (Sec. 26) applies to a reaction 
under conditions of equilibrium. At this point the electrode po- 
tentials of the half -cell equilibria are equal and the over-all poten- 
tial is zero. This gives a method of calculating the numerical value 
of the mass-action equilibrium constant of an oxidation-reduction 
equation from the respective specific electrode potentials. 

EXAMPLE. Calculate the numerical value of the equilibrium 
constant of the reaction 2Fe++ + I 2 ^ 2Fe+++ + 2I~ (which at 
moderate concentrations proceeds only very slightly from left to 
right as written). 
SOLUTION: 

(1) 2Fe++ + 2Fe+++ + 2e 

(2) 21- + I 2 + 2e 

(l)-(2) 2Fe++ + I 2 <=> 2Fe+++ + 21- 

fFe~ K - f l 2 [I~~l 2 

r-n .x-Jorr i ^ K (mass-action constant) 
[Fe++] 2 [I 2 ] 

EI = E% (at equilibrium) 



+0.747 + f log - 0.535 + <"* ,og m 
0.0591 



/[F e ++-h]2 j 2 \ 

log UF^+F " DF?/ = "~* 



2 
0.0591 . [Fe+++] 2 [I-] 2 



log K- -0.212 

log K = -7.0 

K = 1.0 X 10- 7 . Ans. 



88 CALCULATIONS OF ANALYTICAL CHEMISTRY 

Problems 

(See Table VII, Appendix, for the necessary oxidation potentials. Tempera- 
tures are 25C.) 

186. Calculate the potentials of the half cells: 
(a) Hg =* Hg++(0.0010 M) + 2 

(6) Co =* Co+++(0.24 M) + 2e 

(c) Pb ++ (0.050 M) + 2H 2 O ^ PbO 2 + 4H + (0.010 M) + 2e 

(d) H 2 O 2 (0.0020 M) ^O 2 (2 atm.) +2Ff(1.5 M) + 2e 

Ans. (a) +0.77 volt. (6) -0.94 volt, (c) +1.24 volt, (d) +0.78 
volt. 

187. Balance the following equation and express it as the difference between 
two half-cell reactions: 

PbO 2 + Br- + 11+ -> Pb+ + Br 2 + H 2 O. 

Ans. PbO 2 + 2Br- + 4H+ - Pb++ + Br 2 + 211 2 O; (2Br~ & Br 2 + 2e) - 
(Pb+ + + 2H 2 O <* PbO, + 4H+ + 2e). 

188. Write the equation showing the net reaction indicated by each of the 
following pairs of half-cell reactions. Show from the respective electrode 
potentials in which direction each reaction will go, assuming all ion concen- 
trations to be 1 molar, (a) A_g?=>Ag + + e, Cu^Cu ++ + 2e; (6) 1I 2 O 2 <= 
O 2 + 2H-* + 2e, Mn ++ + 2H 2 O ^ MnO 2 + 4H+ + 2e. 

Ans. (a) Cu + 2Ag+ -> Cu ++ + 2Ag; (6) MnO 2 + II 2 O 2 + 2H+ - Mn ++ + 
O 2 + 2H 2 O. 

189. Write complete and balanced equations for the half-cell reactions in- 
dicated by the following changes taking place in acid solution: (a) ^b - > SbO + , 
(6) HNO 2 - NCV, (c) As -^ UAsO 2 , (d) BiO+ -* Bi 2 O 4 , (e) Br 2 -> BrO,- 

Ans. (a) Sb + H 2 o"^ SbO+ + 2H*- + 3e, (ft) HNO 2 + HO ^ NOr + 
3H+ + 2e, (c) As + 2H0 * HAsO, + 311+ + 3e, (tt) 2BiO+ + 2H 2 O ^ 
Bi 2 O 4 + 411 < + 2e, (e) Br 2 + GH 2 O ^ 2BrO.r + 12H+ + 10e. 

190. Write complete and balanced equations for the half-cell reactions 
indicated by the following changes taking place in basic solution: (a) Zn > 
ZnOT, (6) HCHO - HCO 2 - (c) IISnO 2 - -*Sn(OH)r, (d) THs - P, (e) Ag -> 
Ag 2 0. 

Ans. (a) Zn + 4OH- ^ ZnO 2 = + 2H 2 O + 2e, (6) IICIIO + 3OH- ^> 
HCOr + 2II 2 6 + 2e, (c) HSnO 2 - + 3OII- + II 2 O & Sn(OII).,- + 2e, 

(d) PR, + 3011- f P ; + 3H 2 O + 3e, (e) 2Ag + 2OTI- ^> Ag 2 O + II 2 O + 2e. 

191. What is the e.m.f. of the following concentration cell? 



Ag Ag+(0.40 molar) Ag' (0.0010 molar) | Ag 

In what direction is the flow of electrons through the solutions as written? 
Ans. 0.10 volt. Left to right. 



OXIDATION POTENTIALS 



89 



192. What e.m.f. can be obtained at 25C. from the following cell? 

Zn | Zn ++ (0.010 molar) || Ag+(0.30 molar) Ag 

In what direction is the flow of electrons through the solutions? Write an 
equation for the reaction at each electrode and for the net reaction. What 
would the concentration of Ag+ have to be for no current to flow? 

Ans. 1.585 volts. Right to left. Zn -* Zn ++ + 2e, 2Ag f + 2e -> 2Ag, 
Zn + 2Ag' ~>Zii +l + 2Ag. 4.5 X lO" 28 " molar. 

193. Calculate the e.m.f. obtainable from each of the following cells. In 
each case indicate the positive to negative direction of the current in the wire 
connecting the electrodes, and write an equation for the net reaction. 



(a) Cd 
(6) Hg 



Cu +l (1.0M On 



Hg+*-(0.10 M) II Hg f + (0.0071 M) Hg 



(d) Pt 



Ag 



Fe++(0.10 M) 
Fe + ++(0.30 M) 

Mn ff (0.10M) 
MnO 4 -(0.060 M) 



Cr ++ +(0.010 M) 
O 2 Or (0.20) M 
H+(1.0M) 

Su+ + (0.050 M) 



Pt 



II+(0.20 M) 
Ag+Cl (sattl.) 



Sn+^+(0.020 M) 
Fe +l (0.10 M) 



Pt 



Fe+++(0.20 M) 



Pt 



Ans. (a) 0.742 volt, right to left. (6) 0.0340 volt, left to right, (c) 0.56 
volt, right to left, (d) 1.33 volts, left to right, (e) 0.52 volt, right to left. 

194. What e.m.f. can be obtained at 25C. from the following cell? 
Ag | Ag+(0.50 molar) || Cd f+ (1.0 X 10~ 4 molar) Cd 

In what direction do the electrons flow through the solutions as written? 
Write the equation for the reaction taking place at each electrode and for the 
net reaction. What would the concentration of Ag + have to be for no current 
to flow? 

Ans. 1.30 volts. Left to right. 6 X 10~ 23 molar. 

195.. What must be the value of x in order that the following reaction shall 
be at equilibrium? 2A1 + 3Cu++(z molar) ^2A1+++(0.10 M) + 3Cu. 

Ans. 1.6 X 10~ 70 molar. 

196. What must be the value of x for the following reaction to be at equi- 
librium at 25C.? 



Cu + 2Ag+Cc molar) 
Ans. 6 X 10~ 9 molar. 



:Cu++(0.10 molar) + 2Ag 



90 CALCULATIONS OF ANALYTICAL CHEMISTRY 

197. If excess metallic iron is placed in a solution 0.010 molar in Cu ++ , what 
concentration of Cu*" 1 " will remain when equilibrium is reached? The reaction 
is practically complete as follows: Fe + Cu ++ > Fe ++ + Cu. 

Ans. 3 X 10- 29 molar. 

198. When excess metallic zinc is added to a solution 0.010 molar in Ag+, 
what is the theoretical concentration of Ag+ after equilibrium is reached? The 
reaction is practically complete as follows: Zn + 2Ag+ Zn* 4 " + 2Ag. 

Ans. 2 X 10~ 28 . 

199. Equal volumes of a solution 0.10 molar in Fe +f and a solution 0.30 
molar in Ce +++ + are mixed. After the reaction is practically complete (Fe++ + 
Ce++++ > Fe+ + + + Ce+ ++ ) and equilibrium has been attained, what is the 
resulting concentration of Fe ++ ? 

Ans. 6 X 10~ 14 molar. 

200. Calculate the equilibrium constant for each of the following: 

(a) Zn + Cu ++ = Zn++ + Cu 
(6) A1+++ + 3 Ag =* Al + 3Ag+ 

(c) Fe+++ + Ag <=> Fe++ + Ag+ 

(d) 2Ce +++ + I, <= 2Ce ++++ + 21- 

(e) 6F6++ + Cr,07- + 14H+ ^ GFe+++ + 2Ct+++ + 7H 2 O 
(/) 2Fe + + + + 2Br- ^ 2Fe ++ + Br 2 

Ans. (a) 3 X 10 37 . (6) 10' 127 . (c) 0.13. (d) 1.6 X 10~ 31 . (e) 10 56 . 
(/) 1.7 X 10-". 



201. Write complete and balanced equations for the half-cell reactions in- 
dicated by the following changes taking place in acid solution: (a) P > 
H 3 P0 4 , (6) Mn0 2 -> MnOr, (c) U ++++ -> UO 2 ++ , (d) SbO+ -. Sb 2 O 6 , (e) Ch -> 
HC10 2 , (/)S 2 3 -->H 2 S0 3 . 

202. Write complete and balanced equations for the half-cell reactions in- 
dicated by the following changes taking place in basic solution: (a) P > PO 3 ^, 
(6) Pb^-^Pba, (c) S0 3 -->SO 4 -, (d) CrO 2 - -> CrOr, (e) N 2 O 4 -* NO,-, 



203. Calculate the potentials of the following half cells: 

(a) Ag^Ag+(0.010M) + e 

(b) Co < Co ++ (0.063 M) + 2e 

(c) Mn^(0.030 M) + 4H 2 O + MnOr(0.020 M) + 8H+(0.10 M) + 5e 

(d) 2Cr+- H '(0.010 M) + 7II 2 ?=* Cr 2 7 (0.020 M) + 14H + (0.030 M) + 6e 

204. Balance the following equations and express each as the difference 
between two half-cell reactions: (a) Fe + H+ -> Fe+ + + H 2 , (b) Fe ++ -f 
MnOr + H+ -* Fe^-^ 4 - + Mn+ + + H 2 O, (c) Fe + N0 8 - + H^ -> Fe 4 "^ + 
NO + H 2 0, (d) C1 2 + H 2 O - HOC1 + H + + CJ- (e) PbS + H 2 O 2 -> 
PbS0 4 + H 2 O. 



OXIDATION POTENTIALS 91 

205. Balance the following equations and express each as the difference 
between two half-cell reactions: (a) Fe ++ + H 2 O 2 + H+ -> Fe +++ + H 2 O, 
(b) MnOr + H 2 O 2 + 11+ -> Mn++ + O 2 + H 2 O, (c) Cr +++ + MnOr + 
H 2 - Cr 2 O 7 - + Mn++ + H+ (d) I 2 + H 2 O 2 ->H+ + I(V + H 2 O, (e) H 2 SO 8 + 
H 2 S - S + H 2 O. 

206. Write the following as balanced ionic equations and express each as 
the difference between two half-cell reactions: (a) SnSO 4 + K 2 Cr 2 07 + 
H 2 S0 4 - SnCSOdt + K 2 S0 4 + CftCSO*). + H 2 O, (6) As + HNO, + H 2 O - 
H 3 As0 4 + NO, (c) Br 2 + NHOH -> NH 4 Br + N 2 + H 2 O, (d) KI + KIO, + 
HC1 -* I, + H 2 O + KC1. 

207. Write an equation showing the net reaction indicated by each of the 
following pairs of half-cell reactions. Show from the respective electrode 
potentials in which direction each reaction should go, assuming all ion con- 
centrations to be 1 molar, (a) Hg <^ Hg^+ + 2e, Zn *=* Zn++ + 2e; (b) H 2 O 2 <= 
+ 2H+ + 2e, Fe++ ^ Fe++ + + e; (c) Pb++ + 2II 2 O <=* PbO 2 + 4H+ + 
2f, 201- & Cl, + 2e; (d) 2C1~ ^ C1 2 + 2c, 2Br~ ?=* Br 2 + 2e. 

208. Solve the preceding problem with respect to the following half-cell 
reactions: (a) Sn++ v* Sn ++++ + 2e, 2I~ ^ I 2 + 2e; (6) 2O+++ + 7H 2 O <= 
Cr 2 O 7 - + 1411+ + 6e, Fe(CN) 6 s ^ Fe(CN) 6 a + e; (c) Hn^ + 2H 2 O ? 
MnO, + 4H + 2f, Fe ++ <^ Fe ++ + + e; (d) Ag <= Ag+ + e, NO + 2H 2 O ? 
NO 3 - + 411*- + 3e. 

209. The following reactions take place at ordinary concentrations as written: 

y++ -f TiO ++ + 2H+ - V +++ + T1+++ + H 2 O 

Bi + 3Fe +++ + HjO - BiO+ +.3FC++ + 2H+ 

Zn + 2Cr+++-> Zn++ + 2Cr++ 

6Br~ + Cr 2 O 7 - + 14H+ - 3Br 2 + 2Cr +++ + 7II 2 O 

3Ti+++ + BiO+ + 2H 2 O - 3TiO ++ + Bi + 4H<- 

2Fe ++ + NO 3 ~ + 3H+ -> 2Fe +++ + HN0 2 + H 2 

Mg + Zn ++ - Mg ++ + Zn 

C r ++ + V+++ -+ Cr +++ + V++ 

HNO 2 + Br 2 + H 2 O - NO 3 - + 2Br~ + 3H+ 

Convert each of these reactions into two half-cell reactions, placing the re- 
duced form on the left and the oxidized form and electrons on the right. 
Rearrange the 10 half-cell reactions in tabular form in such a manner that the 
strongest reducing agent is at the top left and the strongest oxidizing agent 
is at the bottom right. From the tabulation predict which of the following do 
not react at ordinary concentrations and write balanced ionic equations for 
those that do: (a) Zn++ + V++ (6) Fe++ + TIO++, (c) Cr ++ + N0 8 ~ + 
H+ (d) Bi + Br 2 , (e) Cr 2 O 7 - + TiO+ + + H+. 

210. Which of the following reactions should take place as indicated when 
ion concentrations are 1 molar? (a) 2C1~ + Ii -+ C1 2 + 2I~ (b) 2Fe(CN)- -f 
H 2 O 2 - 2Fe(CN) 6 B + O 2 + 2H + , (c) 2Fe ++ + PbO + 4H+ -> 2Fe +++ + 
Pb ++ + 2H 2 O, (d) 2BJ + 6H+ -> 2Bi +++ + 3H 2 , (e) 1QCT+++ + 6Mn0 4 ~ 4- 
11H 2 O - 5Cr 2 O 7 - + 6Mn ++ + 22H+. 



CALCULATIONS OF ANALYTICAL CHEMISTRY 



211. Calculate the e.m.f. of each of the following cells. In each case 
indicate the direction of the passage of electrons in the wire connecting the 
electrodes, and write an equation for the net reaction. 



Cu 



(a) Ag Ag+(1.0 M) II Cd++(1.0 M) Cd 
(6) Cu | Cu f * (0.010 M) || Cu ' ' (0.090 M) 

Cr Mt (0.050 M) 
Cr 2 O 7 -(0.10 M) 
II + (2.0 M) 



(d) Pt 



Pt 



M) 
Fe +++ (0.0l5 M) 



Sii++(0.020 M) 
Sii+ ++ +(0.080 M) 

Fc ++ (0.25 M) 



1 (0.050 M) 
MnO 4 -(0.10 M) 

II 1 (0.40 M) 

AgCl (siitd.) 



Pt 



Pt 



Fe f M (0.050 M) 

212. What is the c.ni.f. of the following concentration cell? 

Cu | Cu++(1.0 X 10~ 5 molar) || Cu ++ (0.080 molar) Cu 
In what direction is the flow of electrons through the solutions as written? 

213. What e.m.f. can be obtained at 25C. from the following cell? 

Ag Ag + (0.50 molar) || Cd ' f (1.0 X 10- 1 molar) Cd 

In what direction is the flow of electrons through the solutions? Write the 
equation for the reaction at each electrode and for the net reaction. What 
would the concentration of Ag f have to be for no current to llo\v? 

214. A piece of zinc is immersed in a 0.010 molar solution of silver ions at 
25C., and the reaction Zn + 2 Ag ' > Zn ' f + 2Ag is practically complete. 
What is the concentration of silver ions theoretically remaining in the solution? 

215. What must be the value of jc in order that the following reaction shall 
be at equilibrium: Zn + 2Ag f (x molar) ^ Zn + +(1.0 X 10~ 3 molar) + Ag? 

216. When excess metallic aluminum is added to a solution 0.10 molar in 
Cu ++ , what is the theoretical concentration of Cu ++ after equilibrium is 
reached? The reaction is practically complete as follows: 2AJI + 3Cu f+ 
2A1+++ + 3Cu. 

217. Equal volumes of a solution 0.40 molar in Fe+ + and a solution 0.10 
molar in Ce ++ ! f are mixed. After the reaction is practically complete 
(Fe ++ + Ce +++ + > Fe +++ + Ce +++ ) and equilibrium has been attained, 
what is the resulting concentration of Ce^" 1 " 1 " 1 "? 

218. Calculate the equilibrium constant for each of the following: 

(a) CU++ + 2Ag -+ Cu + 2Ag + 

(6) 3Cu ++ + 2A1 ^ 3Cu 

(c) Fe^ + + CG++++ + Fe 

(d) 2Br- + I 2 + Br 2 + 21- 

(e) 101- + 2Mn0 4 ~ + 16II + + 5I 2 + 2Mn + + + 8H 2 



PART II 
GRAVIMETRIC ANALYSIS 

CHAPTER VII 
THE CHEMICAL BALANCE 

47. Sensitiveness of the Chemical Balance. The determina- 
tion of the weight or the mass of a body is a fundamental measure- 
ment of analytical chemistry and is made with an equal-arm 
balance of high degree of precision. An equal-arm balance con- 
sists essentially of a rigid beam supported horizontally at its center 
on a knife-edge and so constructed that the center of gravity of 
the swinging portion is below the point of support. 

The sensitiveness or sensitivity of a chemical balance is the tan- 
gent of the angle through which the equilibrium position of the 
pointer is displaced by a small excess I6ad (usually 1 mg.) on the 
balance pan. This angle is usually so small that it is sufficiently 
accurate to define the sensitiveness of a balance as the number 
of scale divisions through which the equilibrium position of the 
pointer of the balance is displaced by an excess load of 1 mg. 

The sensitiveness varies directly with the length of the balance 
beam, inversely as the weight of the beam, and inversely as the 
distance between the center of gravity of the swinging portion 
and the point of support. That is, 

tan a = ri k 
bd 

where a = the angle through which the pointer is moved 
b = the weight of the beam 
I = the length of the beam 
d = the distance between the center of gravity and the 

point of support 
k = a constant 

93 



94 CALCULATIONS OF ANALYTICAL CHEMISTRY 

The sensitiveness of a balance decreases slightly with increasing 
load. 

48. Method of Swings. In most analytical work, it is suffi- 
ciently accurate to make weighings by the method of short swings. 
The pointer of the balance is allowed to swing only one or two 
scale divisions to the right or left of the zero point of the scale, 
and the reading of the weights is taken when the extreme posi- 
tions of the pointer to the right and left of the zero point are equal. 
The balance is, of course, previously adjusted so that with no 
load on the pans the extreme positions of the pointer are likewise 
equal. 

For more accurate work the method of long swings is used. The 
equilibrium position of the pointer is first determined by allowing 
the beam of the empty balance to swing so that the pointer passes 
over six to eight divisions on the scale. Extreme positions of 
the pointer in an odd number of consecutive swings are recorded; 
for example, three readings are taken to the right of the zero 
point of the scale, and two readings are taken to the left of the 
zero point. The two sets of readings are averaged, and the 
equilibrium position of the pointer is taken as the algebraic mean 
of the two values. As an illustration, assume that the pointer of 
a balance swings as indicated below: 



+4.5 

^-^ 

-3.5 

\_ 

+4.0 



Average: -3.3 +4.1 

Equilibrium position: - 

z 

The weight of an object can then be determined by placing it on 
the left-hand pan and (1) adjusting the weights and rider, or 
weights and chain, so that the equilibrium position is the same 
as that obtained with the empty balance, or (2) calculating the 
weight from the sensitiveness of the balance and the equilibrium 




THE CHEMICAL BALANCE 95 

position corresponding to an approximate weighing of the object. 
For example, suppose the equilibrium position of the pointer under 
zero load is +0.4 as determined above. Suppose that, when an 
object is balanced with 20.1260 grams, the equilibrium position 
is found to be +1.6 and the sensitiveness of the balance under a 
20-gram load is 4.0 (i.e., a 1-mg. increment shifts the equilibrium 
position by 4.0 scale divisions); then the necessary shift of 1.2 
divisions to the left to bring the equilibrium position to +0.4 is 
accomplished by increasing the weight on the right-hand side of 
the balance by 1.2/4.0 = 0.3 mg. The weight of the object is, 
therefore, 20.1263 grams. 

49. Conversion of Weight in Air to Weight in Vacuo. Archi- 
medes 7 principle states that any substance immersed in a fluid 
weighs less by an amount equal to the weight of fluid displaced. 
Consequently, a substance weighed in the ordinary manner is 
buoyed up to a slight extent by the surrounding air, and for 
accurate determinations, especially those involving the weighing 
of objects of large volume, a correction for this buoyant effect 
must be applied. Since the usual method of weighing consists in 
balancing the substance to be weighed against standard weights, 
the surrounding air likewise exerts a buoyant effect upon the 
weights. If the volume occupied by the weights used is equal to 
the volume occupied by the substance, the buoyant effects will 
be equal, and the weight of the substance in vacuo will be the 
same as its weight in air. If the volume occupied by the substance 
is greater than the volume occupied by the weights, the substance 
will weigh more in vacuo than in air; and if the weights have the 
greater volume, the substance will weigh less in vacuo than in air. 
In any case, the difference between the weight in air and the weight 
in vacuo will be equal to the difference between the weight of the 
air displaced by the substance and the weight of the air displaced 
by the weights used. The weight in vacuo W Q may be expressed 
by the equation 



where W = the weight of the substance in air 

V = the volume occupied by the substance 
V = the volume occupied by the weights used 
a = the weight of a unit volume of air 



96 CALCULATIONS OF ANALYTICAL CHEMISTRY 

Since in practice, the volume occupied by the substance and the 
volume occupied by the weights are usually unknown, the formula 
is better written by expressing the values V and V' in terms of 
weight and density. If d is the density of the substance and d' 
the density of the weights, the volume occupied by the substance 
will be W/d, and the volume occupied by the weights used will 
be W/d f . The formula may now be written 



Since the value following the plus sign in this expression is small 
compared with the value W to which it is added and since W 
and W are nearly equal, W may be substituted for the TF in 
the parenthesis without appreciably affecting the accuracy of the 
formula. The formula now becomes 

'W 
W=W + ' 

or 



Although the value of a varies slightly with the temperature 
and barometric pressure, the approximate value of 0.0012 gram 
for the weight of one milliliter of air may be used except in cases 
where extreme accuracy is required or whore the atmospheric con- 
ditions are highly abnormal. The densities of a few common sub- 
stances are shown in Table I. A consideration of the precision of 
the various terms shows that the values of a, d, and d' need be 
known only approximately and that in most cases the computa- 
tion may be performed with sufficient accuracy by means of a 
slide rule. 

TABLE I. DENSITIES OF A FEW COMMON SUBSTANCES 

Aluminum 2.7 Mercury 13.6 

Brass 8.4 Nickel 8.7 

Brass (balance weights) 8.0 Platinum 21 .4 

Copper 8.9 Porcelain 2.4 

Glass 2.6 Quartz 2.7 

Gold 19.3 Silver 10.5 

11.3 Steel 7.8 



THE CHEMICAL BALANCE 97 

EXAMPLE. A platinum crucible weighs 25.0142 grams in air 
against brass analytical weights. What is its weight in vacuo? 
SOLUTION: 

Density of platinum = 21.4 = d 
Density of brass weights = 8.0 = d' 
Weight of 1 ml. of air = 0.0012 gram = a 

Substituting in the above formula, 



- 25.0142 

= 25.0142 - 0.0024 

= 25.0118 grams. Ans. 

50. Calibration of Weights. Tn an ordinary quantitative chem- 
ical analysis, if the same set of weights is used throughout, it is 
immaterial whether or not the masses of the weights are exactly 
as marked so long as they are in correct relative proportion. The 
mass of the 5-gram weight should be exactly one-half that of the 
10-gram weight, and the others should be similarly in proportion. 
In order to establish this relationship and to determine what cor- 
rection factors must be applied to the individual weights of a 
given set, the weights must be calibrated. 

There are several ways of calibrating weights. One of the sim- 
plest is to assume temporarily that one of the smaller weights, 
say tlurlO-mg. weight, is correct and to determine the value of 
the other weights in relation to it. In order to allow for the 
possible fact that the arms of the balance may be slightly unequal 
in length, weighings should be made by the method of substitution, 
which practically eliminates any error from this source. 

Ordinarily the masses of objects A and B are compared by plac- 
ing them on opposite pans of a balance, but by the method of 
substitution, object A is placed on the left-hand pan and is bal- 
anced against a tare (which may be a weight from an auxiliary 
set) on the right-hand pan. A is removed and B is placed on the 
left-hand pan. Tf it exactly balances against the same tare, it 
has the same mass as A. If there is a slight difference, the change 
in position of the rider or chain necessary to reestablish equilib- 
rium can be taken as a measure of the difference in mass. 



98 



CALCULATIONS OF ANALYTICAL CHEMISTRY 



The relative mass of the other 10-mg. weight (which can be 
marked 10') in the set is determined by comparing with the 10-mg. 
weight; the relative mass of the 20-mg. weight is similarly deter- 
mined by balancing against the combined 10-mg. and lO'-mg. 
weights. This process is continued up through the entire set of 
weights, the combined values of the smaller weights being used 

TABLE II. TYPICAL CALIBRATION CORRECTIONS 



Face value of 
weight, 
grams 


True value, 
based on 
0.010-gram 
standard 


Fractional part 
of 10-gram 
standard 


Necessary 
correction, 
mg. 


0.010 


0.0100 


0.0101 


-0.1 


(initial standard) 








0.010' 


0.0101 


0.0101 





0.020 


0.0199 


0.0202 


-0.3 


0.050 


0.0506 


0.0506 





0.100 


0.1012 


0.1011 


+0.1 


0.100' 


0.1014 


0.1015 


-0.1 


0.200 


0.2023 


0.2023 





0.500 


0.5056 


0.5056 





1 


1.0110 


1.0113 


-0.3 


2 


2.0225 


2.0226 


-0.1 


2 


2.0228 


2.0226 


+0.2 


5 


5.0562 


5.0564 


-0.2 


10 


10.1128 


10.1128 





(final standard) 








10' 


10.1130 


10.1128 


+0.2 


20 


20.2262 


20.2256 


+0.6 


50 


50.5635 


50.5640 


-0.5 



to establish the values of the larger weights. In this way values 
similar to those listed in the second column of Table II are ob- 
tained. Because of the small standard taken, it will usually be 
found that the larger weights have large correction factors. It is 
therefore convenient to convert the values to a larger standard, 
say one of the 10-gram weights of the set (or an auxiliary 10-gram 
weight from a set checked by the Bureau of Standards, which has 
been included in the above series of weighings). In the case cited 
in the table, the new 10-gram standard has a value of 10.1128 
grams (relative to the original small standard) . The 5-gram weight 
should have a value of exactly one half of this, or 5.0564 grams. 



THE CHEMICAL BALANCE 99 

Actually its value is 5.0562 grams; hence it is 0.0002 gram too 
light. Therefore 0.2 mg. must be subtracted from a weighing in 
which its face value is used. 

In weighing a given object, instead of applying a correction 
for each weight used, it is less tedious to construct a table showing 
cumulative corrections. By means of such a table the total cor- 
rection is found from the sum of the face values of the weights 
on the pan. In this case it is necessary when weighing an object 
to adopt the convention of using the smallest number of weights 
possible, to use an "unprimed" weight (e.g., 0.100 gram) in pref- 
erence to a " primed" weight (e.g., 0.100' gram), and to construct 
the table accordingly. In the table given, a total weight of 0.18 
gram would show a net correction of -0.3 mg., which is the alge- 
braic sum of the individual corrections of the weights having the 
face values 100, 50, 20, and 10 mg. (i.e., +0.1 + 0.0 - 0.3 - 0.1 = 
-0.3). 

Problems 

219. The addition of a small weight to a certain balance displaces the 
pointer through an angle of 6. Through what angle would a weight \}<-% 
times as great displace it? 

Ans. 8 57'. 

220. A crucible weighing approximately 10 grams is being weighed. The 
pointer of the empty balance has an equilibrium position at +0.2 on the 
scale, and the sensitiveness of the balance under a 10-gram load is known to 
be 3.6 divisions. With the rider at 4.8 (mg.) on the beam, the equilibrium 
point of the balance is found to be at +2.7 on the scale. To what point 
on the beam should the rider be moved to make the correct final reading? 

Ans. 5.5. % 

221. When the pointer of a balance (having a 10-gram load on each pan) is 
set in motion, it swings on the scale as follows: right to +7.6; left to 6.4; 
right to +7.0; left to 5.8; right to +6.2. With an additional 1-mg. weight 
on the right-hand pan, the pointer swings as follows: right to +1.0; left to 
-8.2; right to +0.4; left to -7.6; right to -0.3. What is the sensitiveness 
(in scale divisions) of the balance under a 10-gram load? 

Ans. 4.2 divisions. 

222. The balance of the preceding problem is used to weigh a certain 
crucible and is adjusted so that the equilibrium position of the pointer is at 
zero when the balance is empty. With the crucible on the left-hand pan, with 
weights totalling to 10.12 grams on the right-hand pan, and with the rider at 
3.0 mg. on the right-hand beam, the equilibrium position of the pointer is 



100 CALCULATIONS OF ANALYTICAL CHEMISTRY 

found to be at 1.7 on the scale division. What is the weight of the crucible 
to the nearest tenth of a milligram? 
Ans. 10.1226 grains. 

223. A sample of an alloy having a volume of 2 ml. is weighed in air with 
brass weights and is found to weigh 16.0000 grams. What is its weight in 
vacuo? 

Ans. 16.0000 grams. 

224. Find the weight in vacuo of a piece of gold that weighs 35.0000 grains 
in air against brass weights. 

Ans. 34.9968 grams. 

225. A substance weighing 12.3456 grams in air has a volume of 2 ml. and a 
density equal to three times that of the weights used. What does it weigh 
in vacuo? 

Ans. 12.3408 grams. 

226. A quartz crucible weighing 16.0053 grams in a vacuum would weigh 
how many grams in air against brass weights? 

Ans. 16.0005 grams. 

227. A sample of brass weighs 12.8150 grams in air against platinum 
weights. What is its weight in vacuo? 

Ans. 12.8162 grams. 

228. If a piece of gold in vacuo weighs thirty-five times as much as a 
1-gram brass weight in vacuo, what would the brass weigh in air against 
gold weights? What would the piece of gold weigh in air against brass weights? 

Ans. 0.99991 gram. 35.0031 grams. 

229. In calibrating a given sot of weights the 10-mg. weight is temporarily 
taken as a standard and assumed to be 0.0100 grain. On this basis the 2-gram 
weight is found to be 2.0169 grams and a 10-gram weight (certified by the 
Bureau of Standards to be correct to within less than 0.05 mg.) is found to be 
10.0856 grams. What correction should be applied for the 2-gram weight in 
any weighing in which its face value is taken as its true weight? 

Ans. 0.2 mg. 



230. A weight of 1 mg. on the right-hand pan of a certain balance displaces 
the equilibrium position of the pointer by 6.0 mm. The pointer is 24.6 cm. 
long. What is the tangent of the angle through which the pointer has moved? 
If the beam weighs 32.0 grains and is 16.0 cm. long, what is the distance be- 
tween the middle knife-edge and the center of gravity of the moving parts? 

231. When the pointer of a balance (having a 20-gram scale on each pan) 
is set in motion, it swings on the scale as follows: right to +6.2; left to 6.1 ; 
right to +5.7; left to 5.5; right to +5.3. With an additional 1-iiig. weight 
on the right-hand pan, the pointer swings as follows: right to +3.1; left to 



THE CHEMICAL BALANCE 101 

-9.2; right to +2.6; left to -8.7; right to +2.2. What is the sensitiveness (in 
scale divisions) of the balance under a 20-gram load? 

232. The balance mentioned in the preceding problem is adjusted so that 
the equilibrium position of the pointer of the empty balance is at zero and is 
used to weigh a certain crucible. With the crucible on the left-hand pan, with 
weights adding to 19.87 grams on the right-hand pan, and with the rider at 
8.0 mg. on the right-hand beam, the equilibrium position of the pointer is 
found to be at + 1.2 on the scale division. What is the weight of the crucible 
to the nearest tenth of a milligram? 

233. What would be the weight of a piece of gold in vacuo if in air against 
brass weights it weighs 14.2963 grams? 

234. In vacuo, a quartz dish weighs 22.9632 grams. Calculate the weight 
in air against brass weights. 

235. In determining an atomic weight, a final product which has a density 
of 6.32 is weighed in air against gold weights. What percentage error would 
be made by failing to convert this weight (10.0583 grams) to the weight 
in vacuo? 

236. Find to the nearest tenth of a milligram the weight in vacuo of a 
piece of silver which weighs 20.0113 grams in air against brass weights. 

237. Find the weights in vacuo of two crucibles, one of gold and one of 
aluminum, each weighing 15.0000 grams in air against brass weights. What 
would the gold crucible weigh in air against aluminum weights? 

238. What is the density of a solid which weighs approximately 20 grams 
in air against brass weights and the weight of which increases by exactly 
0.01 per cent in vacuo? What is the density of a similar substance the weight 
of which decreases by 0.01 per cent in vacuo? 

239. From the correction values given in Table II construct a table of cumu- 
lative corrections for weighings ranging from 0.01 to 0.99 gram and construct 
another table of cumulative corrections for weighings ranging from 1.00 gram 
to 99.99 grams. (Hint: To save space use a tabulation similar to that used in 
logarithm tables.) 

240. Assuming that the 20-gram weight in a given set has a value of 20.2364 
grams in relation to the 10-mg. weight as a standard, what should be the value 
of a 500-mg. weight to have a zero correction if the 20-gram weight is taken as 
the final standard and assumed to be 20.0000 grams? If the 500-mg. weight 
actually has a value of 0.5063 on the basis of the smaller standard, what 
correction should be applied for this weight in any weighing in which its face 
value is taken as its true weight? 



CHAPTER VIII 
CALCULATIONS OF GRAVIMETRIC ANALYSIS 

51. Law of Definite Proportions Applied to Calculations of 
Gravimetric Analysis. Gravimetric analysis is based on the law 
of definite proportions, which states that in any pure compound 
the proportions by weight of the constituent elements are always 
the same, and on the law of constancy of composition, which states 
that masses of the elements taking part in a given chemical change 
alw r ays exhibit a definite and invariable ratio to each other. It 
consists in determining the proportionate amount of an element, 
radical, or compound present in a sample by eliminating all in- 
terfering substances and converting the desired constituent or 
component into a weighable compound of definite, known compo- 
sition. Having then determined the weight of this isolated com- 
pound, the weight of the desired component present in the sample 
can be calculated (see also Chap. III). 

EXAMPLE I. A sample of impure sodium chloride is dissolved 
in water, and the chloride is precipitated with silver nitrate ac- 
cording to the following equation. 

NaCl + AgNO 3 -> AgCl + NaNO 3 
or 



furnishing 1.000 gram of silver chloride. What is the weight of 
chlorine in the original sample? 

SOLUTION: Since silver chloride always contains silver and 
chlorine in the respective ratio of their atomic weights, or in the 
ratio of 107.88:35.46, in every 143.34 (107.88 + 35.46) grams of 
silver chloride there are 35.46 grams of chlorine. In 1.000 gram 
of silver chloride there is 

1.000 X ~ = 1-000 X = 0.2474 gram of chlorine. Ans. 



EXAMPLE II. The iron in a sample of FeCOa containing inert 
impurities is converted by solution, oxidation, precipitation, and 

102 



CALCULATIONS OF GRAVIMETRIC ANALYSIS 103 

ignition into Fe 2 3 weighing 1.000 gram. What is the weight of 
iron expressed as FeCO 3 , as Fe, and as FeO in the original sample? 
SOLUTION: The reactions may be expressed by the following 
molecular equations: 

FeCO 3 + 2HC1 -> FeCl 2 + CO 2 + H 2 O 
2FeC0 3 + 4HC1 - 2FeCl 2 + 2CO 2 + 2H 2 
2FeCl 2 + Br 2 + 2HC1 -> 2FeCl 3 + 2HBr 
2FeCl 3 + 6NH 4 OH -> 2Fe(OH) 3 + 6NH 4 C1 
2Fe(OII) 3 -> Fe 2 O 3 + 3H 2 

From these equations it is seen that one molecular weight in 
grams (one mole) of FeC0 3 will furnish one molecular weight in 
grams of FeCl 2 , or 2 moles of FeCO 3 will furnish 2 moles of FeCl 2 . 
Two moles of FeCl 2 will give 2 moles of FeCl 3 , which in turn will 
precipitate 2 moles of Fe(OH) 3 ; and this last compound on igni- 
tion will give one mole of Fe 2 3 . Hence, every 2 moles (231.72 
grams) of FeC0 3 will eventually furnish one mole (159.70) grams 
of Fe 2 O 3 , and it will do so independently of the nature of the 
process or composition of the reagents used to bring about the 
conversion. Indeed, the above reactions are better written in 
the ionic form, thus: 



FeCO 3 + 211+ -> Fe++ + C0 2 + H 2 O 

Br~ 



3NH 4 OII -> Fc(OH) 3 + 3NH 
2Fe(OH)s - Fe^ + 3H 2 O 



+ 



and for purposes of calculations all the intermediate steps may be 
omitted and the fundamental change expressed by the hypo- 
thetical equation 

2FeCO 3 + -> Fe 2 3 + 2CO 2 

In general, it is unnecessary to determine the weights of inter- 
mediate products in a reaction that takes place in steps, and for 
purposes of calculation only the initial and final substances need 
be considered. 

Since two moles (231.72 grams) of FeC0 3 will furnish one mole 
(159.70 grams) of Fe 2 O 3 , by simple proportion, 1.0000 gram of 



9TT* f^O 

Fe 2 3 will be obtained from 1.0000 X 5. ' 3 = 1.0000 X = 

159.70 



104 CALCULATIONS OF ANALYTICAL CHEMISTRY 

1.4510 grams of FeCO 3 . Since each mole of FeC0 3 contains one 
gram-atomic weight (55.85 grams) of Fe and represents the equiva- 
lent of one mole (71.85 grams) of FeO(FeCO 3 -* FeO + C0 2 ), the 
corresponding weights of Fe and FeO would be 



1.0000 X r =1.0000 X 2 * 5 ~ A 85 = 0.6994 gram of Fe, 
1*6203 



and 

9"PVO 9 V 71 R^ 

1.0000 X |^ =1.0000 X 1 ' '.? = 0.8998 gram of FeO, re- 



spectively. Ans. 

EXAMPLE III. What weight of Fe 3 C>4 will furnish 0.5430 gram 
of Fe 2 3 ? 

SOLUTION: Whatever equations may be written to represent 
the conversion of the Fe 3 C>4 to the FeaOa it will be found that from 
every 2 moles of Fe 3 O4 there are obtained 3 moles of Fe 2 O 3 , and 
the hypothetical equation may be written 

2Fe 3 4 + - 3Fe 2 3 
Hence, 

0.5430 X = 0.5430 X 



52. Chemical Factors. A chemical factor may be defined as 
the weight of desired substance equivalent to a unit weight of 
given substance. Thus, in the above three examples, the num- 
bers obtained from the ratios Cl/AgCl, 2FeCO 3 /Fe 2 O 3 , 2Fe/Fe 2 O 3 , 
2FeO/Fe 2 O 3 , and 2Fe 3 C>4/3Fe 2 O 3 are chemical factors since they 
represent the respective weights of Cl, FeCO 3 , Fe, FeO, and Fe 3 C>4 
equivalent to one unit weight of AgCl or of Fe 2 0s as the case 
may be. 

A weight of one substance is said to be equivalent to that of 
another substance when the two will mutually enter into direct 
or indirect reaction in exact respective proportion to those weights. 
In the example cited above, 231.72 grams of FeCO 3 produce 159.70 
grams of Fe 2 O 3 . Hence, 231.72 grams of FeC0 3 are equivalent 
to 159.70 grams of Fe 2 O 3 . The equivalent weights of elements 
and compounds may be expressed by mutual proportions as in 
the case just given, or they may be referred to a common standard 
for which purpose the atomic weight of hydrogen (1.008) is usually 
taken (see Sec. 24). 



CALCULATIONS OF GRAVIMETRIC ANALYSIS 105 

Notice that in expressing a chemical factor the atomic or molec- 
ular weight of the substance sought is placed in the numerator, 
the atomic or molecular weight of the substance weighed is placed 
in the denominator, arid the coefficients are adjusted in accordance 
with the reactions involved. When the principal element, or radical 
desired occurs in both numerator and denominator, usually the 
number of atomic weights of this element or radical will be the 
same in both numerator and denominator, although there are in- 
stances when this is not true. For example, in the reaction 

2CuCl - CuCl 2 + Cu 

the weight of free copper liberated from one gram of cuprous 
chloride is 1.000 X (Cu '2CuCD = 0.3210 gram, and 0.3210 is the 
chemical factor in this particular case. 

That the principal element does not always occur in both nu- 
merator and denominator is shown in the determination of bromine 
by precipitation as silver bromide and conversion to silver chloride 
with a current of chlorine. 

2AgBr + C1 2 -- 2AgCl + Br 2 

Here the weight of bromine represented by one gram of silver 

., . Br 2 2X79.92 AF:r7Q 
chloride is - 2 - - 0.5578 gram. 



53. Calculation of Percentages. Since the chemical factor rep- 
resents the weight of desired element or compound equivalent to 
one unit weight of the element or compound weighed, from any 
weight of the latter the weight of the former can be calculated. 
The percentage of that substance present in the sample may be 
found by dividing by the weight of sample and multiplying by 
100. 

EXAMPLE I. If 2.000 grams of impure sodium chloride are dis- 
solved in water and, with an excess of silver nitrate, 4.6280 grams 
of silver chloride are precipitated, what is the percentage of chlorine 
in the sample? 

SOLUTION: The chemical factor of Cl in AgCl is 0.2474, in- 
dicating that 1 .000 gram of AgCl contains 0.2474 gram of Cl. In 
4.6280 grams of AgCl there are therefore 4.6280 X 0.2474 = 1.145 
grams of Cl. Since this amount represents the chlorine present 



106 CALCULATIONS OF ANALYTICAL CHEMISTRY 

in 2.000 grams of the material, the percentage weight of chlorine 
i 14^ 

must be ^^ X 100 = 57.25 per cent. Ans. 
^.uuu 

EXAMPLE II. A half-gram sample of impure magnetite (Fe 3 4 ) 
is converted by chemical reactions to Fe 2 O 3 , weighing 0.4110 gram. 
What is the percentage of FeaO 4 in the magnetite? 

SOLUTION: The chemical factor of Fe 3 4 from Fe 2 O 3 is 
2Fe 3 O 4 /3Fe 2 03 = 0.9666 which represents the weight of Fe 3 4 
equivalent to 1.000 gram of Fe 2 O 3 . The weight of Fe 3 O 4 equiva- 
lent to 0.4110 gram of Fe 2 3 must be 0.4110 X 0.9666 = 0.3973 

3973 
gram, and the percentage of Fe 3 O 4 in the sample must be 



100 = 79.46 per cent. Ans. 

Problems 

241. Calculate the chemical factors for converting (a) BaSO 4 to Ba, 
(b) Cb 2 6 to Cb, (c) Mg 2 P 2 O 7 to MgO, (d) K01O 4 to K 2 O, (e) Fe 3 O 4 to Fe 2 O 8 . 

Ans. (a) 0.5885, (b) 0.6990, (c) 0.3621, (d) 0.3399, (e) 1.035. 

242. Calculate the chemical factors of the following: 

WEIGHED SOUGHT 

(a) (NII 4 ) 2 PtCl6 NH 3 

(6) MoS 3 MoO 3 

(c) U0 8 U 

(d) B 2 O 3 Na 2 B 4 O 7 .10H 2 O 

(e) (NII 4 ) 3 PO 4 .12Mo0 8 P 2 O 6 

Ans. (a) 0.07671, (6) 0.7492, (c) 0.8480, (d) 2.738, (e) 0.03783. 

243. What is the weight of sulfur in 5.672 grams of barium sulfate? 
Ans. 0.7790 gram. 

244. How many grams of Na 2 SO 4 .10H 2 O are equivalent to the Na in the 
NaCl required to precipitate AgCl from 2.000 grams of AgNO 3 ? 

Ans. 1.896 gram. 

246. A sample "of impure ferrous ammonium sulfate weighs 0.5013 
gram and furnishes 0.0968 gram of Fe 2 O 3 . What is the percentage of 
Fe(NH 4 ) 2 (S0 4 ) 2 .6H 2 0? 

Ans. 94.82 per cent. 

246. A sajnple of limestone weighing 1.2456 grams furnishes 0.0228 gram 
of Fe 2 O 3 , 1.3101 grams of CaSO 4 , and 0.0551 gram of Mg 2 P 2 7 . Calculate 
the percentage of (a) Fe, (6) CaO, (c) MgO in the limestone. What weight 
of C0 2 could be in combination with the calcium? 

Ans. (a) 1.28 per cent, (6) 43.32 per cent, (c) 1.60 per cent. 0.4237 gram. 



CALCULATIONS OF GRAVIMETRIC ANALYSIS 107 

247. What weight of pyrite containing 36.40 per cent of sulfur must have 
been taken for analysis in order to give a precipitate of barium sulfate weighing 
1.0206 grams? 

Ans. 0.3850 gram. 

248. What is the percentage composition of a brass containing only copper, 
lead, and zinc if a half-gram sample furnishes 0.0023 gram of PbSO 4 and 
0.4108 gram of ZnNH 4 PO 4 ? What weight of Zn 2 P 2 O 7 could be obtained by 
igniting the zinc ammonium phosphate? 

Ans. Cu = 69.60 per cent, Pb = 0.31 per cent, Zn = 30.09 per cent. 
0.3510 gram. 

249. What is the percentage of fluorine in a sample of soluble fluoride 
weighing 1.205 grams if it yields a precipitate of CaF 2 weighing 0.4953 gram? 

Ans. 20.01 per cent. 

260. The nitrogen in a half-gram sample of organic material is converted 
to NH 4 HSO 4 by digestion with concentrated H 2 SO 4 . If the NH 4 + ions are 
precipitated as (NH 4 ) 2 PtCl 6 and the precipitate is ignited to Pt, what is the 
percentage of nitrogen in the sample if the metallic platinum weighs 0.1756 
gram? 

Ans. 5.044 per cent. 

261. A sample of pyrite, FeS 2 , contains only silica and other inert impurities 
and weighs 0.5080 gram. After decomposing and dissolving the sample, a 
precipitate of 1.561 gram of BaSO 4 is subsequently obtained. Calculate the 
percentage of sulfur in the sample. If the iron in the solution had been pre- 
cipitated as Fe(OH) 3 and ignited to Fe 2 O 3 , what weight of ignited precipitate 
would have been obtained? 

Ans. 42.20 per cent. 0.2671 gram. 

262. A sample of alum, K 2 SO 4 .A1 2 SO 4 .24H 2 O, containing only inert im- 
purities weighs 1.421 grams. It gives a precipitate of A1(OH) 8 which, after 
ignition to A1 2 O 3 , weighs 0.1410 gram. What is the percentage of S in the 
alum? What is the percentage purity of the alum? 

Ans. 6.246 per cent. 92.43 per cent. 



263. Calculate the chemical factors for converting (a) Fe 2 O 3 to Fe; 
(b) AgCl to KC1O 4 ; (c) Cu3(AsO 3 )2.2As 2 O3.Cu(C 2 H 3 O 2 ) 2 (mol. wt. = 1,014) 
to As 2 3 ; (d) BaSO 4 to HSCN; (e) K 2 A1 2 (SO 4 ) 4 .24H 2 O to H 2 SO 4 . 

264. Calculate the chemical factors of the following: 

WEIGHED SOUGHT 

(a) Mg 2 P 2 7 P 

(6) K 2 PtCl 6 KC1 

(c) Mn3O 4 Mn 2 O 3 

(d) Cu 2 (SCN) 2 HSCN 

(e) KBF 4 Na 2 B 4 O 7 .10H 2 O 



108 CALCULATIONS OF ANALYTICAL CHEMISTRY 

255. What weight of AgBr could be obtained from 4.7527 grams of 
Ag 2 Cr 2 7 ? 

256. How many pounds of phosphorus are contained in 1.000 ton of 

Ca 3 (P0 4 ) 2 ? 

257. A manufacturer using potassium cyanide in a process involving its 
use as cyanide only, substituted sodium cyanide at 45 cents a pound for a 
chemically equivalent quantity of potassium cyanide at $2 a pound. How 
much did he save per pound of KCN? 

268. An ammonium salt is converted into (NH 4 ) 2 PtCL> and the latter 
ignited until only the Ft remains in the crucible. If the residue weighs 
0.1000 gram, what weight of Nils was present in the original salt? 

259. What weight of water could be obtained by strongly igniting 2.000 
grams of datolite [CaB(CII)SiO 4 ] (mol. wt. - 160.0)? 

260. Find the percentage composition of the following in terms of the oxides 
of the metallic elements: (a) FeSO 4 .7H 2 O, (ft) K 2 SO 4 .A1 2 (S0 4 ) 3 .24II 2 O, 
(r) 3Ca 3 (PO 4 ) 2 .CaCO 3 (mol. wt. = 503.7). 

261. An alloy is of the following composition: Cu = 65.40 per cent; 
Pb = 0.24 per cent; Fe = 0.56 per cent; Zn = 33.80 per cent. 

A sample weighing 0.8060 gram is dissolved in HNO 3 and electrolysed. 
Copper is deposited on the cathode; PbO 2 is deposited on the anode. When 
NH 4 OH is added to the residual solution, Fe(OTI) 3 is precipitated and the 
precipitate is ignited to Fe 2 Oa. The zinc in the filtrate is precipitated as 
ZnNH 4 PO 4 and ignited to Zn 2 P 2 7 . What weights of Cu, PbO 2 , Fe 2 O 3 and 
Zn 2 P 2 O 7 were obtained? 

262. How many grams of KNO 3 are equivalent to the potassium iri that 
weight of K 3 PO 4 which contains the same amount of combined P 2 O& that is 
contained in 1.100 grams of Ca 3 (PO 4 ) 2 ? 

263. The antimony in a sample of alloy weighing 0.2500 gram is converted 
to Sb 2 O 6 arid this substance is ignited to Sb 2 O 4 . If the latter weighs 0.1305 
gram, what is the percentage of Sb in the alloy? 

264. A sample of Pb 3 O 4 , containing only inert matter, weighs 0.1753 gram 
and, after dissolving, subsequently yields a precipitate of PbSO 4 weighing 
0.2121 gram. What is the purity of the sample expressed in terms of per- 
centage of Pb? In terms of percentage of Pb 3 O 4 ? 

266. A sample of FeS0 4 .(NII 4 ) 2 S0 4 .6H 2 O containing only inert impurities 
weighs 1.658 grains. After dissolving, oxidizing, and precipitating the iron, 
the Fe(OH) 3 ignites to Fe 2 O 3 and weighs 0.3174 gram. Calculate the per- 
centage of sulfur in the sample. What is the percentage of impurities in the 
sample? 

266. A sample of Mn 2 O 3 is dissolved, and the manganese subsequently 
precipitated as MnO 2 and ignited to Mn 3 O 4 . If the latter weighs 0.1230 gram, 
calculate the weight of Mn 2 3 shown to be present in the original sample. 



CALCULATIONS OF GRAVIMETRIC ANALYSIS 109 

64. Calculation of Atomic Weights. Determinations of atomic- 
weight values at the present time are chiefly revisions of those 
already established, in order that their accuracy may be in keep- 
ing with improved apparatus and methods. In such cases, the 
formulas of the compounds involved are well established and the 
required calculations are thereby made very simple. The experi- 
mental procedure usually followed is to prepare from the element 
a known compound of high degree of purity. This compound is 
weighed, and the percentages of its constituents are determined 
gravimetric-ally. The mathematical computations involved are 
exactly similar to those of an ordinary gravimetric analysis, except 
that the atomic weight of the desired element is the only unknown 
factor. 

EXAMPLE. Carefully purified sodium chloride weighing 2.56823 
grams furnishes 6.2971 grams of silver chloride. Assuming the 
atomic weights of the chlorine and silver to be established as 
35.457 and 107.880, respectively, calculate the atomic weight of 
sodium. 
SOLUTION : 



Weight of NaCl = weight of AgCl X 

NaC1 



2.50823 = G.2971 X 
2.50823 = 0.2971 X 



AgCl 
Na + 35.457 



107.880 + 35.457 
Solving, 

Na = 23.003. Ans. 

Problems 

267. Tf silver phosphate is found by careful analysis to contain 77.300 per 
cent silver, what is the calculated atomic weight of phosphorus (Ag = 107.88)? 

Ans. 31.04. 

268. From an average of 13 experiments, Baxter finds the ratio of silver 
bromide to silver chloride to be 1.310171. If the atomic weight of silver is 
taken as 107.880 and that of chlorine as 35.457, what is the atomic weight of 
bromine? 

Ans. 79.915. 

269. In determining the atomic weight of manganese, Berzelius in 1828 
obtained 0.7225 gram of Mn 2 O 3 from 0.5075 gram of Mn. Von Hauer in 
1857 obtained 13.719 grams of Mn 3 O 4 from 12.7608 grams of MnO. In 1906 



110 CALCULATIONS OF ANALYTICAL CHEMISTRY 

Baxter and Hines obtained an average of 11.43300 grams of AgBr from 
6.53738 grams of MnBr 2 . What are the three values as determined? (Br = 
79.916). 

Ans. 56.66, 55.024, 54.932. 

270. In determining the atomic weight of arsenic, Baxter and Coffin 
converted several samples of Ag 3 As0 4 into AgCl and found the average 
value for the factor 3AgCl/Ag 3 AsO 4 to be 0.929550. Using the factor 
Ag/AgCl as found by Richards and Wells to be 0.752632, calculate the per- 
centage of Ag in Ag 3 AsO 4 . Taking the atomic weight of silver as 107.880, 
calculate to five figures the atomic weight of arsenic. 

Ans. 69.9609 per cent. 74.961. 



271. In determining the atomic weight of aluminum, Richards and Krepelka 
prepared pure samples of AlBrs and experimentally determined the weight of 
silver required to precipitate the halogen. Results of four experiments were 
as follows: 

WT. ALBRJ WT. AG 

Sample 1 5.03798 6.11324 

Sample 2 5.40576 6.55955 

Sample 3 3.41815 4.14786 

Sample 4 1.98012 2.40285 

If the atomic weight of silver is taken as 107.880 and that of bromine is 
taken as 79.916, what is the mean value obtained for the atomic weight of 
aluminum? 

272. The ratio of the weight of silicon tetrachloride to the weight of an 
equivalent amount of silver has been found to be 0.393802 db 0.000008. 
Assuming Cl = 35.457 and Ag = 107.880, calculate the atomic weight of 
silicon. 

273. Classen and Strauch have determined the weights of bismuth oxide 
obtainable from several samples of pure bismuth triphenyl. In one such 
determination, 5.34160 grams of Bi(C c H 6 )3 gave 2.82761 grams of Bi 2 O 3 . 
Calculate the atomic weight of bismuth as shown from these figures. C = 
12.010; H = 1.0080. 

274. What is the atomic weight of copper if 1.0000 gram of Cu is obtained 
by the reduction of 1.2518 grams of CuO? 

276. From the ratios NaNO 3 /NaCl = 1.45422, AgCl/Ag = 1.328668, and 
NaCl/Ag = 0.541854 and assuming N = 14.008, calculate the atomic 
weights of silver, sodium, and chlorine. 

55. Calculations Involving a Factor Weight Sample. It is 
sometimes desirable in industrial work, where large numbers of 
samples of similar material are analyzed, to regulate the weight 
of sample so that the weight of the final product obtained mul- 
tiplied by a simple factor will exactly equal the percentage of the 



CALCULATIONS OF GRAVIMETRIC ANALYSIS 111 

desired constituent. This makes it possible to have the sample 
weighed out directly against a tare, perhaps by someone inexperi- 
enced in exact weighing, and at the same time to eliminate both 
the tedious calculations necessary for each analysis and the pos- 
sibility of mathematical errors. 

The calculation of a desired constituent in a chemical analysis 
involving a direct gravimetric determination is carried out by 
means of the following formula: 

Grams of product X chemical factor ^ 1AA , 

- j ; X 100 = per cent 

Grams ot sample 

Since for a specific determination the chemical factor is a con- 
stant, the expression contains only three variable factors, viz., the 
weight of product, the weight of sample', and the percentage of 
desired constituent. If any two are known, the third can be cal- 
culated; or, since the expression involves only multiplication and 
division, if the numerical ratio between the weight of product and 
the weight of sample, or between the weight of product and the 
percentage of desired constituent, is known, the remaining term 
can be determined. Thus, if the weight of product is numerically 
equal to the percentage of desired constituent, these values cancel, 
and the weight of sample becomes equal to one hundred times 
the chemical factor. If the weight of product is numerically equal 
to the weight of sample, these values cancel, and the percentage 
of desired constituent becomes equal to one hundred times the 
chemical factor. Other ratios may be inserted in the expression, 
and the calculation made in a similar way. 

EXAMPLE. The chemical factor of a certain analysis is 0.3427. 
It is desired to regulate the weight of sample taken so that (a) each 
centigram of the precipitate obtained will represent 1 per cent of 
the desired constituent; (6) every 2 centigrams of precipitate will 
represent 1 per cent of the desired constituent; (c) the percentage 
will be twice the number of centigrams of precipitate; (rf) three- 
fourths of the weight in grams of precipitate will be one-fiftieth 
of the percentage of desired constituent. What weight of sample 
should be taken in each case? 

SOLUTION: (a) The relation between the weight of precipitate 
and the percentage of constituent is such that 0.01 gram =s= 1 per 
cent. Hence, 



112 CALCULATIONS OF ANALYTICAL CHEMISTRY 
0.01 X 0.3427 



X 

x = 0.3427 gram. A ns. 
OJ2X 0.3427 X100 _, 

X 

x = O.G854 gram. Ans. 
(c) 0.01 X0.3427 x 1QO _ 2 

# = 0.1714 gram. 



a; = 0.9135 gram. Ans. 

Problems 

276. In the analysis of a sample of feldspar for silica, the sample is fused, 
dissolved in HC1, and the solution is evaporated to dryness, heated, and 
treated with acid. The residue is filtered off and weighed as SiO 2 . What 
weight of sample should he taken for analysis so that (a) each centigram of 
residue will represent 1 per cent SiO 2 , (b) the number of centigrams will 
represent directly the percentage SiO 2 , (r) every 2 centigrams of residue will 
represent 1 per cent SiO-2, (d) twice the number of centigrams will equal the 
percentage of Si02? 

Ans. (a) 1.000 gram, (fe) 1.000 gram, (r) 2.000 grams, (d) 0.5000 gram. 

277. What weight of cast iron should be taken for analysis so that the 
weight of ignited SiO in centigrams will be equal to one-third of the per- 
centage of Si in the cast iron? 

A?is. 0.156 gram. 

278. Calculate the weight of limestone to be taken so that the number of 
centigrams of CaO obtained and the percentage of Ca in the sample will be 
in the respective ratio of 7:5. 

Ans. 1.001 grams. 

279. In the analysis of a sample of lead salt, the lead is determined as 
PbSO 4 . What weight of sample should be taken for analysis such that the 
weight of lead sulfate in grams obtained will represent (a) one-thirtieth the 
percentage of lead in the sample expressed as Pb, (b) one-thirtieth the per- 
centage of lead expressed as PbO 2 , (c) one-thirtieth the percentage of lead 
expressed as PbCrO 4 ? 

Ans. (a) 2.277 grams, (6) 2.629 grams, (r) 3.553 grams. 

280. A sample of ammonium salt is analyzed by precipitating the ammo- 
nium radical as (NH^Ptdfl and igniting the precipitate to metallic platinum. 
Calculate the weight of sample to be taken for analysis (a) so that the weight 



CALCULATIONS OF GRAVIMETRIC ANALYSIS 113 

of ignited precipitate in grams multiplied by the atomic weight of platinum 
will give the percentage of N in the sample, (b) so that the weight of ignited 
precipitate in milligrams multiplied by ^ioo will equal four-thirds of the per- 
centage of NH 3 in the sample. 

Ans. (a) 0.07357 gram, (b) 0.7753 gram. 



281. What weight of impure ferrous sulfate should be taken for analysis 
so that each milligram of Fe 2 O 3 obtained will correspond to 0.120 per cent FeO 
in the sample? 

282. What weight of dolomite should be taken for analysis so that, in the 
determination of magnesium, the number of centigrams of Mg 2 P2O 7 obtained 
will be twice the percentage of Mg in the mineral? 

283. In the analysis of potassium in a silicate, the mineral is decomposed 
and the potassium subsequently weighed as KCKXj. What weight of sample 
was taken if it was found that the percentage of K 2 O in the mineral could 
be found by dividing the number of milligrams of KC1O 4 obtained by 12? 

284. What weight of magnetite (impure Fe a O 4 ) should be taken for 
analysis so that after decomposition of the sample, precipitation of the iron 
as Fe(OTT) 3 , and ignition to Fe 2 Oj (n) the number of centigrams of FcoO,* 
obtained will be equal to the percentage of Fe.s() 4 in the sample, (/>) the number 
of milligrams of Fc 2 O 3 obtained will be live times the percentage of Fe 3 O 4 , 
(r) the percentage of Fc in the sample and the number of centigrams of 
Fe 2 O 3 obtained will be in the ratio of 3:2? 

56. Calculation of the Volume of a Reagent Required for a 
Given Reaction. The volume of a solution required to carry out 
a given reaction can be calculated if the concentration of the solu- 
tion is known. If the concentration is expressed in terms of nor- 
mality, the calculation is best made by the methods of volumetric 
analysis, i.e., in terms of milliequivalents (see Sec. 21); if the con- 
cent-ration is expressed as grams of solute per unit volume of solu- 
tion or in terms of specific gravity and percentage composition, 
the calculation is usually easiest made by the use of the chemical 
factor. 

EXAMPLE I. How many milliliters of barium chloride solution 
containing 90.0 grams of liuCl 2 .2H 2 per liter are required 
to precipitate the sulfate as BaSO 4 from 10.0 grams of pure 
Na 2 $CVH)H 2 0? 

SOLUTION: The weight of BaCl 2 .2H 2 O for the precipitation is 
found by means of the chemical factor, thus: 

10.0 X J^HrfiT^ 10 - X -Q99 = 7 - 58 8 rams of BaCl 2 .2H 2 
JNa2olJ4.1vUl2vJ o^ 



114 CALCULATIONS OF ANALYTICAL CHEMISTRY 

Since each milliliter of reagent contains 0.0900 gram of BaCl2.2H 2 O, 
the volume of solution required is 

7.58 



0.0900 



= 84.2 ml. Ans. 



When the concentration of the required reagent is expressed 
in terms of the percentage by weight of the solute, the specific 
gravity of the solution must also be known in order to determine 
the volume required. As stated in Sec. 21, there is no exact 
mathematical relationship between these two factors, but tables 
are given in all standard chemical handbooks showing this rela- 
tionship for solutions of common substances experimentally deter- 
mined at many different concentrations. Consequently, when a 
problem includes only one of these factors, tables must be con- 
sulted in order to determine the other. In the Appendix, specific- 
gravity-percentage tables are given for a few common acids and 
bases. These tables apply to weighings in vacuo at definite tem- 
peratures, but since three-significant figure accuracy is all that is 
needed in most calculations involving specific gravity of solutions, 
it is usually not necessary to make corrections for temperature 
and buoyancy differences. 

EXAMPLE II. How many milliliters of ammonia water of specific 
gravity 0.950 (containing 12.72 per cent of NH 3 by weight) are 
required to precipitate the iron from 0.800 gram of pure ferrous 
ammonium sulfate, FeSO4.(NH 4 ) 2 S04.6H 2 0, after oxidation of the 
iron to the ferric state? 

SOLUTION: Since 3 molecules of ammonia are required to pre- 
cipitate one atom of ferric iron 

Fe+++ + 3NII 3 + 3H 2 O -> Fe(OH) 3 + 3NH 4 + 

it follows that the weight of NH 3 necessary to precipitate the 
iron from 0.800 gram of ferrous ammonium sulfate will be 

3NH * 51 - 10 



FeS0 4 .(NH4) 2 S0 4 .6H 2 " ' 392.1 " 

0. 1043 gram of NH* 

Since the ammonia water has a specific gravity of 0.950 and con- 
tains 12.74 per cent of NH 3 by weight, one milliliter of the solu- 
tion weighs 0.950 gram of which 12.74 per cent by weight is NH 3 



CALCULATIONS OF GRAVIMETRIC ANALYSIS 115 

and 87.26 per cent by weight is water. The actual weight of NH 3 
in one milliliter of solution is therefore 0.950X0.1274 = 0.121 
gram." Since 0.1043 gram of NH 3 is required to precipitate the 
iron and since each milliliter of the solution contains 0.121 gram 
of NH 3 , it follows that the volume of solution required is 

0.1043 



0.121 



= 0.862 ml. Ans. 



As explained in Sec. 51 in calculations of this type, the com- 
putations should not be carried through unnecessary steps. In 
the example above, it is not necessary to compute the weight of 
iron contained in the ferrous ammonium sulfate, the weight of 
ammonium hydroxide required to precipitate the iron, and the 
weight of anhydrous ammonia contained in the ammonium hy- 
droxide. On expressing the whole, the common factors cancel. 

0800 
UU 



FeSO4.(NH 4 )2SO4.6H 2 

QXTTT- 

_= 0.1043 gram of NHs 



3NH 3 



(3NII40II) 
0.1043 



0.121 



= 0.862 ml. 



In general, with problems of this type, time will be saved if the 
final multiplications and divisions are not made until all the factors 
are combined and expressed as a whole. In the above example 
the only essential factors are 

0.800X51.10 ,, QA o i A 

= 0.862 ml. Ans. 



392.1X0.950X0.1274 

A very similar type of problem is one in which it is required to 
calculate the volume of a solution of given percentage composi- 
tion required to react with a certain volume of another solution 
of given percentage composition. By computing the weight of 
reacting component in the given volume of the latter solution, 
the problem becomes exactly like the one discussed above. 

EXAMPLE III. How many milliliters of sulfuric acid (sp. gr. 
1.135, containing 18.96 per cent H^SOi by weight) are required 
to neutralize 75.0 ml. of ammonium hydroxide (sp. gr. 0.960, con- 
taining 9.91 per cent NH 3 by weight)? 



116 CALCULATIONS OF ANALYTICAL CHEMISTRY 

SOLUTION: In 75.0 ml. of the ammonia solution there are 

75.0 X 0.960 X 0.0991 grams of NH 3 
The required weight of HC1 for this NH 3 is 

H 2 S0 4 



75.0 X 0.960 X 0.0991 X 



2NH 3 

QQ no 

75.0 X 0.960 X 0.0991 X grams 



Since each milliliter of the acid contains 1.140 X 0.2766 gram of 
HC1, the volume of acid required is 

75.0 X 0.960 X 0.0991 X 98.08 OK , , 
1.135 X 0.1896 X 34.06 = 95 ' 6 ml An8 ' 

Problems 

285. What volume of ammonium oxalate solution [35.1 grams of 
(NH4) 2 C 2 O4.Il2O per liter] will be required to precipitate the calcium as 
CaC 2 O 4 from 0.124 gram of 3Ca 3 (PO 4 )2.CaCl 2 ? What volume of "magnesia 
mixture" containing 1 F.W. of MgCl 2 per liter will be necessary to precipitate 
the phosphate as MgNH 4 PO 4 from the filtrate from the calcium determina- 
tion? 

Ans. 4.82 ml. 0.71 ml. 

286. How many milliliters of silver nitrate solution containing 20.00 grams 
of AgNO 3 per 100 ml. are required to precipitate all the chloride as AgCl from 
a solution containing 2.012 grams of dissolved BaCl 2 .2H 2 O? How many milli- 
liters of H 2 S() 4 (sp. gr. 1.105) are required to precipitate the barium as BaSO 4 
from the same solution? 

Ans. 14.00ml. 4.86ml. 

287. In the precipitation of arsenic as MgNH 4 AsO 4 from a solution of 
0.4000 gram of pure As 2 Oa that has been oxidized to arsenic acid, it is desired 
to add sufficient magnesium chloride reagent (64.00 grams MgCl 2 per liter) to 
precipitate the arsenic and also have 200 mg. of Mg remaining in solution. 
What volume is required? 

Ans. 18.26 ml. 

288. Chloride samples are to be prepared for student analysis by using the 
chlorides of sodium, potassium, and ammonium, alone or mixed in various 
proportions. How many milliliters of 5.00 per cent silver nitrate of specific 
gravity 1.041 must be added to a 0.300-gram sample in order to ensure com- 
plete precipitation in every possible case? 

Ans. 18.3 ml. 



CALCULATIONS OF GRAVIMETRIC ANALYSIS 117 

289. How many milliliters of aqueous ammonia (sp. gr. 0.900) are re- 
quired to precipitate the iron as Fe(OH) 8 from a half-gram sample of pure 
Fe 2 (S0 4 ) 3 .9II 2 O? 

Ans. 0.36 ml. 

290. Calculate the volume of hydrochloric acid (sp. gr. 1.050, containing 
10.17 per cent HC1 by weight) to neutralize (a) 48.6 ml. of a solution of 
KOH (sp. gr. 1.100, containing 12.0 per cent KOH by weight), (6) 152.1 ml. 
of a solution of NaOH (sp. gr. 1.327), (c) a solution containing 10.0 grams of 
pure KOH, (d) a solution containing 10.0 grams of impure KOH (96.6 per 
cent KOH, 2.2 per cent K 2 CO 3 , 1.2 per cent H 2 O), (e) 25.3 ml. of ammonia 
water containing 15.04 per cent by weight of NH 3 . 

Ans. (a) 39.0 ml., (6) 508 ml, (c) 60.9 ml., (d) 59.8 ml., (e) 72.0 ml. 

291. The following are added to water: 1.60 grams of pure Na 2 CO 3 , 2.21 ml. 
of H 2 SO 4 solution (sp. gr. 1.700), and 16.0 ml. of KOH solution (56.0 grams of 
solid per liter). This solution is to be brought to the exact neutral point. 
The solutions available for this purpose are hydrochloric acid (sp. gr. 1.141) 
and ammonia water (sp. gr. 0.930). Which should be used? What volume is 
required? 

Ans. 1.26 ml. of ammonia water. 

292. In the reaction 2NaCl + H 2 SO 4 -> Na 2 SO 4 + 2HC1, it is desired to 
add sufficient sulfuric acid (sp. gr. 1.835) to liberate that amount of HC1 
which when absorbed in water will furnish 250 ml. of solution of specific 
gravity 1.040. Calculate the volume necessary. 

Ans. 16.7 ml. 

293. A solution of ferrous ammonium sulfate is prepared by dissolving 
2.200 grams of pure FeSO 4 .(NII 4 ) 2 SO 4 .6H 2 O in 500 ml. of water containing 
15.0 ml. H 2 SO 4 (sp. gr. 1.135, containing 18.96 per cent H 2 S() 4 by weight). 
The iron is then oxidized by bromine water (2Fe+ + + Br 2 > 2Fe+++ + 2Br~). 
What total volume of NH 4 OII (sp. gr. 0.950, containing 12.74 per cent NH 3 by 
weight) is required to neutralize the acid and just precipitate all of the iron as 
Fe(OH) 3 ? 

Ans. 11.6 ml. 

294. How many milliliters of barium chloride solution containing 21.05 grams 
of BaCl 2 .2H 2 O per liter are required to precipitate all the sulfate as BaSO 4 from 
a solution containing 1.500 grams of dissolved Fe 2 (SO 4 ) 8 .9H 2 O? How many 
milliliters of NaOH solution (sp. gr. 1.200) are required to precipitate all the 
iron as Fe(OH) 3 from the same solution? 

295. A sample of MgCO 3 contaminated with SiO 2 weighs 0.5000 gram and 
loses 0.1002 gram on ignition to MgO. What volume of disodium phosphate 
solution (90.0 grams Na 2 HPO 4 .12H 2 O per liter) will be required to precipitate 
the magnesium as MgNH 4 PO 4 ? 

296. How many milliliters of a solution of potassium dichromate contain- 
ing 26.30 grams of K 2 Cr 2 O 7 per liter must be taken in order to yield 0.6033 
gram of Cr 2 O 3 after reduction, precipitation, and ignition of the chromium? 



118 CALCULATIONS OF ANALYTICAL CHEMISTRY 



297. The arsenic in a half-gram sample of As2Sa is oxidized to arsenic 
acid, and is precipitated with a solution of " magnesia mixture" (MgCU + 
NI^Cl). If exactly 12.6 ml. of the mixture are required, how many grams of 
MgCl 2 per liter does the solution contain? (H 3 AsO 4 + MgCl 2 + 3NH 4 OH > 
MgNH4AsO 4 + 2NH 4 C1 + 3H 2 O.) 

298. How many grams of silver chloride will be formed by the addition 
of an excess of silver nitrate to 10.00 ml. of hydrochloric acid (sp. gr. 1.160, 
containing 31.52 per cent HC1 by weight)? 

299. Sulfuric acid of specific gravity 1.800 is to be used to precipitate the 
barium as barium sulfate from 1.242 grams of pure BaCl 2 .2H 2 O. Calculate 
the volume of acid necessary for precipitation. 

300. How many milliliters of ammonia (sp. gr. 0.940) will neutralize 
40.00 ml. of H 2 SO 4 solution (sp. gr. 1.240)? 

301. According to the following equation what volume of HNOs (sp. gr. 
1.050) is required to oxidize the iron in 1.000 gram of FeS0 4 .7II 2 in the 
presence of sulfuric acid? [6FeSO 4 + 2IINO 3 + 3H 2 SO 4 --> 3Fe 2 (SO 4 ) 3 + 
2NO + 4TI 2 O.] 

302. How many milliliters of ammonia (sp. gr. 0.960 containing 9.91 
per cent of NH 3 by weight) will be required to precipitate the aluminum as 
A1(OH) 3 from a solution containing 500 grams of alum [KA1(SO 4 )2.12II 2 O) 
and 100 ml. of HC1 (sp. gr. 1.120, containing 23.82 per cent HC1 by weight)? 

303. Alum, KA1(SO 4 ) 2 .12H 2 O, weighing 0.6000 gram is dissolved in water 
and 10.0 ml. of hydrochloric; acid (sp. gr. 1.120, containing 23.82 per cent HC1 
by weight) are added. It takes 5.11 ml. of ammonia (containing 28.33 per 
cent NH 8 by weight) to neutralize the acid and precipitate the aluminum 
as A1(OH)3. Find the specific gravity of the ammonia, the normality of the 
acid, and the ratio of the base to the acid. 

304. To a suspension of 0.310 gram of A1(OH) 3 in water are added 13.0 ml. 
of aqueous ammonia (sp. gr. 0.900). How many milliliters of sulfuric acid 
(sp. gr. 1.18) must be added to the mixture in order to bring the aluminum 
into solution? 

57. Indirect Analyses. Problems relating to indirect analyses 
differ from those of direct analyses in much the same way that 
algebraic equations involving two or more unknown quantities dif- 
fer from those involving only one unknown quantity. Indeed, 
such indirect problems are often solved by algebraic methods. 

The simplest type of indirect problem is that in which two pure 
chemical substances are isolated and weighed together. Then 
either by further chemical action on the substances or by chemi- 
cal analysis of a new sample of the same material, additional data 
are derived by which one of the components is determined. The 
other component is then found by difference. 



CALCULATIONS OF GRAVIMETRIC ANALYSIS 119 

In any case, results of analyses of this type are usually less 
precise than results of analyses in which a single component is 
determined by a direct method. In solving simultaneous alge- 
braic equations, for example, there is often a decrease in the 
number of significant figures that may properly be retained. For 
example, in solving the following simultaneous equations: 

0.2395z + 0.2689y = 1.937 
0.2067X + 0.2689y = 1.222 
0.0328z = 0.715 

x = 2.18 

there is a decrease from four-significant- to three-significant-figure 
precision. 

EXAMPLE I. In the analysis of a two-gram sample of lime- 
stone, the weight of combined oxides of iron and aluminum 
(Fe 2 O 3 + A1 2 O 3 ) is found to be 0.0812 gram. By volumetric 
methods, the percentage of total iron calculated as FeO is found 
to be 1.50 per cent. What is the percentage of A1 2 3 in the 
sample? 
SOLUTION : 

i rjn 

Weight of FeO = 2.00 X -~ = 0.0300 gram 



Weight of Fe 2 O 3 = 0.0300 X ~~ = 0.0333 gram 



Weight of A1 2 O 3 = 0.0812 - 0.0333 = 0.0479 gram 

0479 

Percentage of A1 2 O 3 = ' nn X 100 = 2.40 per cent. Ans. 

^.uu 

A second general type of indirect analysis is that in which two 
chemical substances are isolated and weighed together. Then an- 
other measure of the two substances is obtained either by con- 
verting them to two different compounds and again finding the 
combined weights or by determining the amount of reagent 
required to effect such conversion. In this way, by the use of 
algebraic symbols to represent the unknown quantities, two inde- 
pendent equations can be formulated, and from them the values of 
the unknowns can be determined. It is evident that this type of 
problem may be extended to any number of unknown quantities, 



120 CALCULATIONS OF ANALYTICAL CHEMISTRY 

provided sufficient data are given to allow the formulation of as 
many independent algebraic equations as there are unknowns. 

EXAMPLE II. In the analysis of a 0.5000-gram sample of feld- 
spar, a mixture of the chlorides of sodium and potassium is ob- 
tained which weighs 0.1180 gram. Subsequent treatment with 
silver nitrate furnishes 0.2451 gram of silver chloride. What is 
the percentage of Na 2 O and K 2 O in the sample? 

SOLUTION: 

Let x = weight of KC1 

y = weight of NaCl 
(1) Z + i/ = 0.1 180 

Number of grams of AgCl obtainable from x grams of KC1 



Number of grams of AgCl obtainable from y grams of NaCl 



= 2.452ty 
Therefore, 

(2) 1.923* + 2.452?y = 0.2451 

Solving (1) and (2) simultaneously, 

x = 0.0837 gram of KC1 
y = 0.0343 gram of NaCl 

T> , fT-n /K 2 \00837X 100 lftr 
Percentage of K 2 = (s T ) ^rSnn = 10 - G 



, yd -MO 

P 4. fAi o i'NaO\0.0343X 100 ~ f . 
Percentage of NazO = { OM^T) Q^QQ ' = 3 - 64 



The standard J. Lawrence Smith method for determining sodium 
and potassium in a silicate is an example of an indirect analysis 
of this type. By this method the sample is decomposed and the 
alkalies are isolated and weighed as combined chlorides. These 
are dissolved and the potassium alone is precipitated from water- 
alcohol solution as K^PtCle (or as KC1O4) and weighed as such. 
The K 2 PtCl 6 can also be ignited and weighed as 2KC1 + Pt, or 
(after washing) as metallic platinum (see also Part VI). 



CALCULATIONS OF GRAVIMETRIC ANALYSIS 121 

EXAMPLE III. In the analysis of a sample of feldspar weighing 
0.4150 gram, a mixture of KC1 + NaCl is obtained weighing 0.0715 
gram. From these chlorides 0.1548 gram of K 2 PtCl6 is obtained. 
Calculate the percentage of Na 2 O in the feldspar. 
SOLUTION: 

Let x = weight of NaCl in combined chlorides 
Then 

0.0715 - x = weight of KC1 



(0.0715 - x) x 6 = 0.1548 

/IVL/l 

Solving, 

x = 0.0415 gram 

0.0415 X (Na 2 O/2NaCl) vx inn c OA . XT ~ , 

- n \ 1 * ' - - X 100 = 5.30 per cent Na 2 O. Ans. 
0.4150 

Problems 

306. A mixture of 0.2600 gram of ferric oxide and 0.4500 gram of alumi- 
num oxide is ignited in hydrogen, the ferric oxide alone being reduced to 
metallic iron. What is the final weight? 

Ans. 0.6318 gram. 

306. A silicate weighing 0.6000 gram yields a mixture of pure NaCl and 
pure KOI weighing 0.1800 gram. In this residue the KC1 is converted to 
K 2 PtCl 6 weighing 0.2700 gram. Find the percentage of K 2 O and the percentage 
of Na 2 O in the silicate. 

A ns. 8.72 per cent K 2 O, 8.60 per cent Na 2 O. 

307. In the analysis of a sample of feldspar weighing 0.7500 gram there is 
obtained 0.2200 gram of NaCl + KC1. These chlorides are dissolved in 
water-alcohol mixture and treated with chloroplatinic acid. The precipitate 
of K 2 PtCl 6 is filtered on a Gooch crucible, dried, arid ignited in hydrogen. 
After washing with hot water, the residual platinum then weighs 0.0950 gram. 
Compute the percentages of Na 2 O and K 2 O in the feldspar. What weight of 
precipitate would have been obtained if perchloric acid had been used as the 
precipitating agent and the precipitate had been dried and weighed without 
ignition? 

Ans. 10.42 per cent Na 2 0, 6.11 per cent K 2 O. 0.1349 gram. 

308. A mixture of silver chloride and silver bromide is found to contain 
66.35 per cent of silver. What is the percentage of bromine? 

Ans. 21.3 per cent. 

309. A sample of carbonate rock weighing 1.250 grams yields a precipitate 
of the hydrated oxides of iron and aluminum. These are filtered off and ignited. 
The combined oxides Fe 2 3 + A1 2 3 are found to weigh 0.1175 gram. Iron is 



122 CALCULATIONS OF ANALYTICAL CHEMISTRY 

determined by a volumetric method on a separate sample of the rock and the 
results show 3.22 per cent Fe. Calculate the percentage of Al in the rock. 
Ans. 2.54 per cent. 

310. An alloy weighing 0.2500 gram when treated with HNO 3 gives a residue 
of the hydrated oxides of tin and antimony. These on ignition yield 0.1260 
gram of the combined oxides SnO 2 + Sb 2 C>4. This residue is brought into 
solution and is found by a volumetric method to contain 32.56 per cent Sn. 
Calculate the percentage of antimony in the original alloy. 

Ans. 23 A per cent. 

311. A mixture of silver chloride and silver bromide weighs 0.5267 gram. 
By treatment with chlorine, the silver bromide is converted into silver chloride, 
and the total weight of silver chloride becomes 0.4269 gram. What is the 
weight of bromine in the original mixture? 

Ans. 0.179 gram. 

312. A mixture of pure CaO and pure BaO weighing 0.6411 gram yields 
1.1201 grams of pure mixed sulfates. Find the percentages of Ba and of 
Ca in the original mixture. 

Ans. 17.6 per cent Ca, 67.4 per cent Ba. 

313. A sample of silicate weighing 0.6000 gram yields 0.1803 gram of a 
mixture of pure NaCl and pure KC1. When these are dissolved and treated 
with AgNO 3 , the resulting precipitate of AgCl is found to weigh 0.3904 gram. 
Calculate the percentages of Na 2 O and K 2 O. 

Ans. 7.32 per cent Na 2 0, 10.27 per cent K 2 O. 

314. From a sample of feldspar a mixture of KC1 and NaCl is obtained 
that weighs 0.1506 gram and contains 55.00 per cent chlorine. What weight 
of K 2 PtCl 6 could be obtained from the KC1? 

Ans. 0.212 gram. 

315. A mixture of BaCl 2 .2II 2 O and LiCl weighs 0.6000 gram and with 
silver nitrate solution yields 1.440 grams AgCl. Calculate the percentage of 
Ba in the original mixture. 

Ans. 25.0 per cent. 

316. A mixture of pure NaCl and pure Nal weighs 0.4000 gram and yields 
with AgNO 3 a precipitate of AgCl and Agl that weighs 0.8981 gram. Find 
the percentage of iodine present in the original mixture. 

Ans. 19.8 per cent. 

317. What percentage of MgCO 3 is present with pure BaCO 8 so that the 
mixture contains the same CO 2 content as if it were pure CaCO 3 ? 

Ans. 72.5 per cent. 



318. A 1.0045-gram sample containing only CaCO 3 and MgCO 3 is strongly 
ignited. The weight of the ignited product (CaO and MgO) is 0.5184 gram. 



CALCULATIONS OF GRAVIMETRIC ANALYSIS 123 

Calculate the percentages of Ca and of Mg in the original sample and in the 
ignited sample. 

319. A mixture of BaO and CaO weighing 1.792 grams, when treated with 
sulfuric acid and transformed to mixed sulfates, weighs twice the original 
amount. What is the percentage of BaO in the mixture? 

320. Iron and aluminum are precipitated from a sample of a mineral weigh- 
ing 0.9505 gram and the combined oxides A1 2 O 3 + Fe2O 3 are found to weigh 
0.1083 gram. By a volumetric method this oxide residue is found to contain 
10.50 per cent Fe. What is the percentage of Al in the original mineral? 

321. An alloy weighing 0.5180 gram yields a residue of the hydrated oxides 
of tin and antimony which on ignition produces 0.1661 gram of the combined 
oxides SnC>2 + Sb2O4. By a titration method a separate sample of the alloy 
shows the presence of 10.12 per cent Sb. Calculate the percentage of Sri in 
the alloy. 

322. A silicate rock weighing 0.7410 gram is analyzed by the J. L. Smith 
method, and a mixture of the chlorides of sodium and potassium weighing 
0.2172 gram is obtained. These chlorides are dissolved in a mixture of alcohol 
and water and treated with I1C1O 4 . The dried precipitate of KC1O 4 weighs 
0.3330 gram. What is the percentage of Na2O in the silicate? If the potassium 
had been precipitated as K 2 PtCl6 and the precipitate converted to metallic 
platinum, what weight of platinum would have been obtained? 

323. How many grams of BaCO 3 must be added to 2.40 grams of MgCOs 
so that the mixture will contain the same percentage of CO 2 as CaCO 3 does? 

324. A mixture of NaBr, Nal, and NaNO 3 weighs 0.6500 gram. With 
silver nitrate, a precipitate of the halides of silver is obtained and is found 
to weigh 0.9390 gram. When heated in a current of chlorine gas the pre- 
cipitate is converted entirely into AgCl weighing 0.6566 gram. What is the 
percentage composition of the original sample? 

325. A precipitate of AgOl + AgBr weighs 0.8132 gram. On heating in 
a current of chlorine, the AgBr is converted into AgCl, the mixture losing 
0.1450 gram in weight. What was the percentage of chlorine in the original 
precipitate? 

326. A one-gram sample consisting of a mixture of LiCl, KC1, NaCl, and 
NH 4 C1 is dissolved in water and the solution is divided into two equal parts. 
One portion gives with chloroplatinic acid in the presence of alcohol a pre- 
cipitate of K^PtCle that weighs 0.9500 gram and that loses 35.00 per cent of 
its weight on ignition. The other portion gives a precipitate with AgNO 3 
weighing 1.2500 grams. What is the percentage of each constituent in the 
mixture? 

327. A mixture of NH 4 C1 and KC1 weighs 0.5000 gram. With chloro- 
platinic acid a precipitate is obtained that, when ignited, weighs 1.0400 grams. 
What is the percentage of NH 3 in the mixture? If the ignited precipitate were 
washed with water and reignited, what would be the weight obtained? 

328. A mixture of AgCl and AgBr contains chlorine and bromine in the 
proportion by weight of Cl:Br = 1:2. What is the percentage of silver in 



124 CALCULATIONS OF ANALYTICAL CHEMISTRY 

the mixture? If one gram of the sample were heated in a current of chlorine, 
thereby converting the AgBr into AgCl, what would be the weight of the 
resulting mixture? 

329. A sample of an impure mixture of NaCl, NaBr, and Nal weighing 
1.5000 grams is dissolved in water, arid the solution is divided into two equal 
portions. One portion gives a precipitate of Pdl2 weighing 0.1103 gram. 
The other portion gives a precipitate of AgCl + AgBr + Agl weighing 
1.2312 grams; and when these salts are heated in a current of chlorine they 
are all converted into AgCl weighing 1.0500 grams. What are the percentages 
of NaCl, NaBr, and Nal in the original sample? 

330. A mixture of silver bromide and silver iodide weighs x grams. After 
the mixture has been heated in a current of chlorine, the resulting silver 
chloride is found to weigh y grams. Derive an expression for the percentages 
of bromine and of iodine in the original mixture. 

331. A mixture of silver chloride and silver iodide on being heated in a 
current of chlorine is converted entirely into silver chloride and is found to 
have lost 6.00 per cent of its weight. What is the percentage of chlorine in the 
original mixture? 



CHAPTER IX 
ELECTROLYTIC METHODS 

68. Decomposition Potential. The decomposition potential of 
an electrolyte is the lowest e.m.f. that must be applied in order 
to bring about continuous decomposition of cation and anion at 
the electrodes. 

If a nitric or sulfuric acid solution of copper is electrolyzed be- 
tween platinum electrodes, the copper plates out on the cathode. 

CU++ + 2e -> Cu 
Water is decomposed at the anode. 



As a result there is produced a voltaic cell of the type 
Cu Cu++ 2H+ 2 Pt 



This cell exerts a "back e.m.f." which opposes the applied voltage 
and which can be calculated from the formula 

log [Cu+ 



#H 2 o + log [H+] 2 [press. 



( 



In order to continue the electrolysis, a voltage at least equal to 
this must be applied. In addition, enough voltage to overcome 
the simple ohmic resistance of the solution (E = IK) is necessary, 
and, in cases where polarization effects occur, the required e.m.f. 
must be still further increased. 

Decomposition potential = back e.m.f. + IR + overvoltage 

The extent of overvoltage depends on several factors, such as 
current density (amperes per square centimeter of electrode sur- 
face), concentration, temperature, nature of the substances liber- 
ated, and the character of the electrodes. 

125 



126 CALCULATIONS OF ANALYTICAL CHEMISTRY 

During an electrolysis, those ions in solution will first be dis- 
charged which have the lowest decomposition potentials. So long 
as these ions are present around the electrode in considerable con- 
centration, they, almost alone, are discharged; but, as their con- 
centration diminishes, other ions whose deposition potentials are 
higher but still within that of the current applied will also begin 
to be discharged. 

69. Analysis by Electrolysis. Quantitative analysis by means 
of electrochemical methods is usually restricted to the determina- 
tion of metals. An electric current is passed, under suitable con- 
ditions, through a solution of the salt of a metal, and the metal 
itself is gradually deposited, usually in the elementary condition, 
upon one of the electrodes. The calculation of the amount of 
metal which will be deposited at the end of a given time is founded 
on Faraday's laws which may be stated as follows: 

1. The mass of any substance deposited at an electrode is propor- 
tional to the quantity of electricity which passes through the solution. 

2. The amounts of different substances liberated at the electrodes 
by the passage of the same quantity of electricity are proportional to 
the equivalent weights of the substances. 

Current strength is expressed in terms of the ampere, which 
is defined as that strength of current which, when passed through 
a solution of silver nitrate under certain standard conditions, will 
deposit silver at the rate of 0.001118 gram per second. 

Quantities of electricity are expressed in terms of the coulomb, 
which is defined as that quantity of electricity which passes through 
a conductor in 1 second when the current is 1 ampere. That is, 

Q = It 

where Q = quantity of electricity, coulombs 
/ = current strength, amperes 
t = time, seconds 

From Faraday's first law, it follows that the weight of a sub- 
stance liberated from solution by electrolysis during a given time 
will be directly proportional to the current strength and under a 
given amperage will be directly proportional to the time. 

Faraday's second law states that the weights of different sub- 
stances liberated at the electrodes by a given quantity of elec- 
tricity are proportional to the respective equivalent weights. The 



ELECTROLYTIC METHODS 127 

equivalent weight of a substance in this case is the atomic or 
molecular weight divided by the total change in oxidation number. 
It is found by experiment that 90,500 coulombs are required tc 
liberate a gram-equivalent weight (equivalent weight in grams) of 
any substance. Thus, 96,500 coulombs of electricity are capable 
of depositing at the cathode: 

Ag 

= 107.88 grams of silver from a solution of a silver salt 

Fe 

-75- = 27.92 grams of iron from a solution of a ferrous salt 

Zi 

Fe 

-5- = 18.61 grams of iron from a solution of a ferric salt 

o 

The value 96,500 coulombs is therefore a unit of quantity in 
electrochemical measurements, and in that capacity it is called 
a faraday. One faraday = 96,500 coulombs = 96,500 ampere- 
seconds = 26.81 ampere-hours. 

The above reactions may be expressed by equations, as follows: 

(a) Ag + + 6->Ag 



(c) 

where the symbol represents the electron, or unit of negative 
electricity. If the equations are considered as representing gram- 
atomic or gram-molecular ratios, then the symbol represents the 
faraday. That is, 1, 2, and 3 faradays are required to deposit a 
gram-atomic weight of metal from a solution of silver salt, ferrous 
salt, and ferric salt, respectively. 

EXAMPLE I. How many grams of copper will be deposited in 
3.00 hours by a current of 4.00 amperes, on the assumption that 
no other reactions take place at the cathode? 
SOLUTION: 

t = 3.00 X 3,600 = 10,800 seconds 

Number of coulombs = It = 4.00 X 10,800 = 43,200 



1 faraday would deposit ~ = 31.8 grams of copper 

& 

40 onrj 
43,200 coulombs would deposit X 31.8 = 

14.2 grams Cu. Ans. 



128 CALCULATIONS OF ANALYTICAL CHEMISTRY 

Faraday's laws apply to each electrode. A current of 1 ampere 
flowing for 96,500 seconds through a copper sulfate solution is 
not only capable of depositing Cu/2 = 31.79 grams of copper at 
the cathode, but at the same time will liberate O/2 = 8.000 grams 
of oxygen at the anode. 

Cu++ + 2* -> Cu (cathode) 

H 2 -> 2H+ + HO 2 + 2e (anode) 



The passage of 2 faradays therefore deposits one gram-atomic 
weight of copper (63.57 grams) and liberates ^ mole, or 16 grams, 
of oxygen gas, occupying 1L.2 liters under standard conditions. 
As seen from the equation, the acidity of the solution increases 
by one gram equivalent (=0= 1 mole HNO 3 == 1 A ^ole H 2 SO 4 , etc.) 
per faraday passed. This gain in acidity may be expressed in 
terms of acid normality (see Sees. 24 and 68). It is possible to 
prepare a solution suitable as a standard in acidimetry by elec- 
trolyzing a neutral solution of copper sulfate and determining the 
acid concentration of the resulting solution from the weight of 
copper deposited. 

The electrolysis of a dilute sulf uric acid solution causes the fol- 
lowing reactions to take place : 

2H+ + 2e -> H 2 (cathode) 

H 2 -> 2H+ + H0 2 + 2 (anode) 



The passage of 1 faraday liberates Yz mole of hydrogen (1.008 
grams; 11.2 liters under standard conditions) and Y mole of 
oxygen (8.000 grams; 5.6 liters under standard conditions). The 
total amount of gas evolved per faraday is therefore % mole 
(16.8 liters). In this electrolysis, there is no net change in acidity if 
the volume of the solution is kept constant, for the loss in acidity 
at the cathode is balanced by the gain in acidity at the anode. 

The electrolysis of a dilute nitric acid solution of lead causes 
the deposition of lead dioxide on the anode: 

Pb++ + 2H 2 - PbO 2 + 4H+ + 2e (anode) 
2H+ +2e -+H 2 (cathode) 

One faraday therefore deposits PbO 2 /2 = 119.6 grams of PbO 2 , 
liberates Yi mole of hydrogen, and increases the acidity by 1 gram- 
equivalent weight (1.008 grams) of hydrogen ion. 



ELECTROLYTIC METHODS 129 

EXAMPLE II. A neutral solution containing 0.4000 gram of 
copper is electrolyzed until all the copper is plated out, and the 
electrolysis is continued 7 minutes longer. The volume of the 
solution is kept at 100 ml., and the current strength is maintained 
at an average of 1.20 amperes. On the basis of 100 per cent cur- 
rent efficiency (a) how long did it take for the copper to deposit, 
(6) what total volume of gas was evolved during the entire elec- 
trolysis, (c) what was the acidity of the solution at the end of the 
electrolysis? 
SOLUTION: 

1 faraday deposits -^- = 31.8 grams Cu 
z 

f\ Af\f\f\ 

Number of faradays to deposit 0.4000 gram Cu = HpTo"" == 0.0126 

ul.o 

m. - , 0.0126X96,500 t mo , 

Time required = - T~w\ - = 1*012 seconds = 

16.9 minutes. Ans. 

During deposition of Cu, each faraday (i.e., each ^ mole Cu) 
corresponds to J^ mole O 2 . 

Moles O 2 evolved = 0.0126 X M = 0.00315 
After Cu deposited, number faradays passed = 
7 X 60 X 1.2 



96,500 
Each faraday evolves J/ mole H 2 and % mole O 2 

Moles H 2 + O 2 evolved - 0.00522 X % = 0.00392 
Total gas evolved = 0.00315 + 0.00392 = 0.00707 mole. Ans. 

During Cu deposition each faraday corresponds to a gain of 
1 gram atom of H+ . 

After Cu deposition, acidity does not change 

Gain in acidity = 0.0126 gram-equivalent of H+. Ans. 

(Resulting solution is 10 X 0.01258 = 0.1258 normal as an acid.) 

In the above calculations involving Faraday's second law, it 
has been assumed that all the current serves for the decomposi- 
tion of the substance in question; i.e., 100 per cent current effi- 



130 CALCULATIONS OF ANALYTICAL CHEMISTRY 

ciency has been assumed. In actual analyses, this is not usually 
the case. The electrolysis of an acid solution of a copper salt will 
not only cause the deposition of copper at the cathode, but small 
amounts of hydrogen will usually be given off at the same elec- 
trode before the copper has all plated out. In such cases, the sum 
of the weights of the products discharged at each electrode exactly 
corresponds to the law. That is, in the copper electrolysis, for 
each faraday of electricity passed, the number of gram-equivalent 
weights of copper deposited added to the number of gram-equiva- 
lent weights of hydrogen liberated will be unity. In problems of 
electroanalyses, unless otherwise specified, 100 per cent current 
efficiency may be assumed. 

Other electrical units which are frequently used in electro- 
chemical computations are as follows : 

The ohm, R, is the unit of resistance. It is the resistance offered 
to a constant current of electricity at 0C. by a column of mercury 
1 sq. mm. in cross section and 106.3 cm. long. 

The voltj E, is the unit of electromotive force or electrical pres- 
sure. Its relation to the ampere and ohm is expressed by Ohm's 
law. 



The joule, J, is the unit of work. It is represented by the energy 
expended in 1 second by a current of 1 ampere against a resistance 

of 1 ohm. 

J = Elt - EQ - 10 7 ergs 

The watt, W, is the unit of power. It is represented by the work 
done at the rate of 1 joule per second. 



J = 

Problems 

332. A 100-watt 110- volt incandescent lamp is connected in series with an 
electrolytic cell. What weight of cadmium could be deposited from solution 
by the current in 30 minutes? 

Ans. 0.953 gram. 

333. How many minutes will it take for a current of 0.500 ampere to cause 
the deposition of 500 mg. of silver from nitric acid solution on the basis of 
80 per cent current efficiency? 

Ans. 18.6 minutes. 



ELECTROLYTIC METHODS 131 

334. How many coulombs of electricity are required to deposit 0.1000 
gram of cobalt from a solution of cobaltous salt? How many amperes would 
be required to deposit that amount in 20 minutes 20 seconds? How many 
grams of nickel would be deposited under identical conditions? 

Ans. 327.3 coulombs. 0.2682 ampere. 0.09951 gram. 

336. With a current at 8.00 volts, how much electrical energy is theoretically 
required to deposit (a) 0.100 gram of gold, (6) 0.100 gram of mercury from 
solutions containing these metals in the higher state of oxidation? 

Ans. (a) 1,180 joules, (6) 770 joules. 

336. What weights of Cu, of Zn, and of PbO 2 will be deposited in separate 
electrolytic cells, on the assumption of 100 per cent current yield by a current 
of 0.0800 ampere flowing for 30 hours? 

Ans. 2.85 grams Cu, 2.93 grams Zn, 10.7 grams PbO 2 . 

337. Using a rotating electrode, Sand obtained 0.240 gram of copper from 
a nitric acid solution of copper sulfate in 6 minutes. A current of 10.0 amperes 
under 2.80 volts was used. What electrical energy was expended and what 
was the current efficiency? 

Ans. 10,080 joules, 20.2 per cent. 

338. Using a rotating electrode, Langness found that with a current of 
17.0 amperes and a potential of 10.0 volts, 0.200 gram of platinum could be 
deposited in 5 minutes from a solution of potassium chloroplatinate. How 
much electrical energy was expended per second? What quantity of electricity 
was used? What was the current efficiency? 

Ans. 170 joules. 5,100 coulombs. 7.75 per cent. 

339. What quantity of electricity is required for (a) the electrolytic deposi- 
tion of 1.196 grams of PbO 2 , (6) the liberation of 0.800 gram of oxygen gas, 
(c) the liberation of 30.0 ml. of chlorine when measured under standard condi- 
tions (one molecular weight in grams occupies 22.4 liters)? 

Aris. (a) 965 coulombs, (6) 9,650 coulombs, (c) 259 coulombs. 

340. For how long a time must a current of 1.00 ampere be passed through 
a dilute solution of sulfuric acid in order to liberate a total volume of 600 
ml. of gas when measured dry and under standard conditions? 

Ans. 57 A minutes. 

341. With a current of 1.00 ampere, what weight of silver would be deposited 
in 1 minute in a silver coulometer? What volume of gas (under standard 
conditions) would be evolved in 60 seconds in a water coulometer? 

Ans. 0.06708 gram. 10.44 ml. 

342. Pure crystals of copper sulfate are dissolved in water, and the solution 
is electrolyzed until the solution is colorless. The cathode gains 0.4280 gram. 
What is the acidity of the solution in terms of moles of H 2 SO 4 ? 

Ans. 0.006731 mole. 



132 CALCULATIONS OF ANALYTICAL CHEMISTRY 

343. What would be the net gain or loss in gram-equivalents of H+ per 
faraday in the electrolysis of a solution of HNOs in which 80 per cent of the 
current goes to the simple decomposition of water and 20 per cent goes to the 
reaction involving the reduction of nitrate to free nitrogen at the cathode and 
liberation of oxygen at the anode? 

Ans. 0.040 gram-equivalent lost. 

344. An alloy consists of 20.72 per cent lead and 79.30 per cent zinc. A 
1-gram sample is dissolved in acid and the solution diluted to exactly one liter. 
A 100-ml. pipetful is titrated with 0.1000 N NaOH and requires 30.00 ml. to 
neutralize the acid present. The remaining 900 ml. are electrolyzed under 
2.00 amperes for exactly 5 minutes (100 per cent current efficiency) with the 
deposition of the lead as PbO 2 . (a) How many milliliters of gas (standard 
conditions) are evolved during this time? (ft) If the volume of the solution 
after electrolysis is brought again to 900 ml., what would be the acid normality 
of the solution. 

Ans. (a) 94.4 ml. (6) 0.034 N. 

346. An average current of 0.5000 ampere is passed through a dilute acid 
solution of an alloy containing 0.5000 gram of copper, 0.2000 gram of zinc, and 
0.1000 gram of lead. Assuming 100 per cent efficiency, compute the total gain 
in acidity in terms of moles of H 2 SO 4 (a) at the end of the PbO 2 deposition at 
the anode, (6) at the end of the Cu deposition at the cathode, (c) after the cur- 
rent has been continued 5 minutes longer. What volume of gas (standard 
conditions) has been evolved during the entire process? 

Ans. (a) 0.0009652 mole, (h) 0.008347 mole, (c) 0.008347 mole. 108.8 ml. 



346. How many milligrams of silver can be deposited from solution in 23 
minutes by a current of 0.700 ampere at 100 per cent efficiency? How long 
would it take the same current to deposit the same weight of nickel? 

347. If a current of 0.250 ampere is passed through a copper sulfate solu- 
tion and a 90 per cent current yield is obtained, compute (a) the weight of 
copper which can be deposited in one hour, (6) the gain in acidity in moles of 
sulfuric acid during that time. 

348. If 6.30 amperes will deposit 0.532 gram of tin in 20 minutes from a 
solution of stannous salt, what is the current efficiency? 

349. A pure alloy of copper and zinc is dissolved in acid and electrolysed 
under 0.500 ampere (100 per cent efficiency). It is found that just 40 minutes 
are required to deposit all the copper. From the filtrate the zinc is precipi- 
tated and ignited to 0.245 gram of Zn 2 P2O 7 . Calculate (a) grams of copper 
deposited, (b) percentage of copper in the alloy, (c) volume of gas liberated, 
(d) gain in acid normality of the solution assuming the volume at the end 
of the electrolysis to be 500 ml. 

350. Assuming the volume of an electrolytic cell to be kept at 125 ml., 
find the normal acid concentration of a solution containing 0.42 gram of 
copper, 0.20 gram of lead, and 0.26 gram of zinc and 144 milliequivalents 



ELECTROLYTIC METHODS 133 

of acid at the start (a) after just enough current has been passed to deposit 
all the lead as PbO 2 and assuming no other reaction takes place at the anode 
during that time, (6) after all the copper has deposited, (c) when the current 
of 1.8 amperes has been continued 30 minutes longer. Compute the total 
volume of gas evolved during the entire electrolysis, measured under standard 
conditions. 

351. On the basis of 30 per cent current yield, compute the cost of the 
power required to produce 1 pound of NaMnO 4 from a manganese anode and 
0.3 N Na 2 CO 3 solution. E.m.f. = 8 volts. Cost of current = 3 cents per 
kilowatt-hour. 

352. The following represents the net reaction for the discharge of an 
ordinary lead storage battery: Pb + PbO 2 + 2H+ + 2HSO 4 - - 2PbSO 4 + 
2H 2 O. Write the two half-cell reactions involved. If the charged battery 
contains two liters of sulfuric acid of specific gravity 1.28 (36.9 per cent H 2 SO 4 
by weight) and in 1 hour the cell is discharged to the point where the specific 
gravity of the acid is 1.11 (15.7 per cent H 2 SO 4 by weight), what is the average 
amperage produced? Assume no change in volume of the electrolyte. 

353. A current of one ampere is passed for one hour through a saturated 
solution of sodium chloride connected in series with a copper coulometer con- 
sisting of copper electrodes dipping in a solution of cupric sulfate. Write 
equations for the anode and cathode reactions that take place in the NaCl 
cell and in the CuSO 4 cell. Assuming 100 per cent current efficiency, calculate 
(a) grams of copper deposited on the cathode of the coulometer, (6) number of 
liters of C1 2 gas evolved (standard conditions), (c) number of milliliters of 
N/10 HC1 required to titrate one-tenth of the cathode portion of the NaCl cell. 

354. A copper coulometer, consisting of copper electrodes dipping in a 
solution of cupric sulfate, is connected in series with a cell containing a con- 
centrated solution of sodium chloride. After a direct current is passed through 
both the cell and the coulometer for exactly 50 minutes, it is found that 0.636 
gram of copper has been dissolved from the coulometer anode, (a) What is 
the average amperage used during the electrolysis? (6) Assuming 100 per 
cent efficiency, how many grams of NaClO 3 could be produced from the C1 2 
and NaOII formed during the electrolysis (3C1 2 + 6OH~ -> C1O 3 - + 5C1- + 
3H 2 0)? 

355. What weight of CuSO 4 .5H 2 O must be dissolved in water so that after 
complete deposition of the copper by electrolysis, a solution will be obtained 
which is equivalent to 100 ml. of 0.100 N acid? 

356. Pure crystals of CuSO 4 .5H 2 O weighing 1.0000 gram are dissolved in 
water, and the solution is electrolyzed with an average current of 1.30 amperes 
for 20 minutes. What weight of copper has been deposited? What volume 
of gas measured dry and under standard conditions has been liberated? If 
the resulting solution is made up to 100 ml. with water, what is its normality 
as an acid? 

357. Pure crystals of CuSO 4 .5H 2 O are dissolved in water, and the solution 
is electrolyzed with an average current of 0.600 ampere. The electrolysis is 



134 CALCULATIONS OF ANALYTICAL CHEMISTRY 

continued for 5 minutes after all the copper has been deposited, and it is 
found that a total volume of 62.5 ml. of gas measured dry at 18C. and 
745 mm. pressure has been evolved. What weight of crystals was taken for 
electrolysis? (Assume all the copper is deposited before hydrogen is evolved.) 
How many milliliters of 0.100 N NaOH will the resulting solution neutralize? 

368. A solution of brass in nitric acid contains 1.10 grams of copper and 
0.50 gram of zinc and is 2.00 N in acid. It is electrolyzed at 1.50 amperes, and 
the volume is kept at 100 ml. If all the current goes to the deposition of 
copper at the cathode, what is the acid normality of the solution when all the 
copper has just deposited? What is the acid normality of the solution if the 
current is continued 20 minutes longer and 40 per cent of it goes to the reduc- 
tion of nitrate ions to ammonium ions: (cathode) NOa" + 8 + 10H+ = 
NH 4 + + 3H 2 0; (anode) 4II 2 O - 8 = 8H+ + 20 2 . How long before the acid 
would be entirely destroyed? 

359. Assuming 100 per cent current efficiency and assuming that the elec- 
trolysis is discontinued as soon as the copper is deposited, compute the time 
required, the volume of gas evolved, and the gain in acidity in terms of 
millimoles of hydrogen ions when 0.8000 gram of brass in dilute HNOs 
is electrolyzed. Cathode gains 0.6365 gram; anode gains 0.0240 gram. 
Current = 0.900 ampere. Compute also the percentage composition of the 
brass. 



CHAPTER X 
CALCULATIONS FROM REPORTED PERCENTAGES 

60. Calculations Involving the Elimination or Introduction of 
a Constituent. It is occasionally necessary to eliminate from or 
introduce into a report of an analysis one or more constituents, 
and calculate the results to a new basis. Thus, a mineral may 
contain hygroscopic water which is not an integral part of the 
molecular structure. After complete analysis, it may be desir- 
able to calculate the results to a dry basis as being more repre- 
sentative of the mineral under normal conditions. On the other 
hand, a material may contain a very large amount of water, and 
because of the difficulty of proper sampling, a small sample may 
be taken for the determination of the water while the bulk of the 
material is dried, sampled, and analyzed. It may then be desir- 
able to convert the results thus obtained to the basis of the original 
wet sample. This applies equally well to constituents other than 
water, and, in any case, the method by which these calculations 
are made is based upon the fact that the constituents other than 
the ones eliminated or introduced are all changed in the same 
proportion, and the total percentage must remain the same. 

EXAMPLE I. A sample of lime gave the following analysis: 

CaO = 90.15 per cent 

MgO = 6.14 per cent 

Fe 2 O 3 + A1 2 O 3 = 1.03 per cent 

SiO 2 = 0.55 per cent 

H 2 O + CO 2 (by loss on ignition) = 2.16 per cent 

100.03 per cent 

What is the percentage composition of the ignited sample on the 
assumption that the volatile constituents are completely expelled? 
SOLUTION: In the sample as given, total percentage of all 
constituents is 100.03. The slight variation from the theoretical 
100 per cent is due to experimental errors in the analysis. The 

135 



136 



CALCULATIONS OF ANALYTICAL CHEMISTRY 



total percentage of nonvolatile constituents is 100.03 2.16 = 
97.87 per cent. Ignition of the sample would therefore increase 
the percentage of each of the nonvolatile constituents in the ratio 
of 100.03:97.87, and the percentage composition of the ignited 
sample would be 

100.03 



CaO 



MgO 



SiO 2 



= 90.15 X 



= 6.14 X 



= 1.03 X 



= 0.55 X 



97.87 

100.03 

97.87 

100.03 

97.87 

100.03 

97.87 



92.14 per cent 
6.28 per cent 
1.05 per cent 
0.56 per cent 



Ans. 



100.03 per cent 



EXAMPLE II. If the original sample of lime mentioned in the 
preceding problem were heated only sufficiently to reduce the 
percentage of volatile constituents from 2.16 to 0.50 per cent, 
what would be the percentage composition of the product? 

SOLUTION: In the original sample, total percentage of non- 
volatile constituents is 100.03 - 2.16 = 97.87 per cent. In the 
ignited sample, the total percentage of residual constituents would 
be 100.03 - 0.50 = 99.53 per cent. The loss of volatile matter 
would therefore have caused the percentage of the various con- 
stituents to increase in the ratio of 99.53 to 97.87. Hence, the 
percentage composition would be 



CaO 
MgO 



= 90.15 X 
= 6.14 X 



97.87 
99.53 
97.87 



= 91.68 per cent 
= 6.24 per cent 



Fe 2 O 3 + Al 2 3 = 1.03XJ~|= 1.05 per cent 

99 53 
= 0.55 X T^T^ = 0.56 per cent 



SiO 2 

Volatile matter 



97.87 



= 0.50 per cent 
100.03 per cent 



Ans. 



61. Cases Where Simultaneous Volatilization and Oxidation or 
Reduction Occur. Occasionally a material on ignition may not 



CALCULATIONS FROM REPORTED PERCENTAGES 137 

only lose volatile constituents but may also undergo changes due 
to oxidation or reduction effects. In such cases the percentages 
of the constituents after ignition can best be calculated by assum- 
ing that oxidation or reduction occurs first, and loss of volatile 
material afterward. In other words, it is easiest to solve the prob- 
lem in two separate steps. 
EXAMPLE. A mineral analyzes as follows: 

CaO = 45.18 per cent 

MgO = 8.10 per cent 

FeO = 4.00 per cent 

SiO 2 = 6.02 per cent 

CO 2 = 34.07 per cent 

H 2 O = 2.03 per cent 
100.00 per cent 

After heating in oxygen the ignited material shows the presence 
of no water and 3.30 per cent CO 2 . The iron is all oxidized to 
the ferric state. Calculate the percentage of CaO and of Fe 2 O 3 in 
the ignited material. 

SOLUTION: Assume first that 100 grams of the original mineral 
are taken and that the only change is that of oxidation of FeO to 
Fe 2 O 3 . Here 4.00 grams of FeO would form 4.00 X Fe 2 3 /2FeO = 
4.44 grams of Fe 2 3 and the resulting material would gain in 
weight by 0.44 gram due to this change alone. The material now 
weighs 100.44 grams and the percentages of the constituents (other 
than Fe 2 O 3 ) are decreased in the ratio of 100/100.44. The per- 
centages are now 

CaO = 45.18 X 100/100.44 = 44.98 per cent 

MgO = 8.10 X 100/100.44 = 8.07 per cent 

FeO = 4.44 X 100/100.44 = 4.42 per cent 

SiO 2 = 6.02 X 100/100.44 = 5.99 per cent 

CO 2 = 34.67 X 100/100.44 = 34.52 per cent 

H 2 = 2.03 X 100/100.44 = 2.02 per cent 

100.00 per cent 

Assume now that the second change takes place, namely, that 
all the water is lost and that the percentage of C0 2 in the ignited 
material is brought down to 3.30 per cent due to loss of most of 



138 CALCULATIONS OF ANALYTICAL CHEMISTRY 

the C02. The calculation then becomes similar to that of the 
preceding example: 

100 00 3 30 
Percentage of CaO = 44.98 X mo( ) _' (34.52 + 2 .02) 

= 60.97 per cent. Ans. 

f ** A , 100.00-3.30 
Percentage of Fe 2 O 3 = 4.44 X 100 . 0() - (34 .52 + 2.02) 

= 6.61 per cent. Ans. 

Problems 

360. The percentage of copper in a sample of copper ore with a moisture 
content of 8.27 per cent is found to be 36.47 per cent. Calculate the percentage 
on a dry sample. 

Ans. 39.76 per cent. 

361. A sample of coal as taken from the mine contains 8.32 per cent ash. 
An air-dried sample of the same coal contains 10.03 per cent ash and 0.53 per 
cent moisture. Calculate the percentage of moisture in the original sample. 

Ans. 17.50 per cent. 

362. A powder consisting of a mixture of pure BaCl 2 .2H 2 O and silica con- 
tains 20.50 per cent Cl. What would be the percentage of Ba in the material 
after all the water of crystallization is expelled by ignition? 

Ans. 44.34 per cent. 

363. A sample of lime gives the following analysis: 

CaO = 75.12 per cent 
MgO = 15.81 per cent 

Si0 2 = 2.13 per cent 
Fe 2 O 3 = 1.60 per cent 

CO 2 = 2.16 per cent 

H 2 O = 3 JL4 per cent 
99.96 per cent 

What is the percentage of each constituent after superficial heating in which 
the CO 2 content has been reduced to 1.08 per cent and the water content to 
1.00 per cent? 

Ans. CaO = 77.66, MgO = 16.35, SiO 2 = 2.21, Fe 2 O 3 = 1.66, CO 2 = 1.08, 
H 2 O = 1.00. 

364. Lime is to be manufactured by the ignition of a sample of dolomite. 
The only data as to the composition of the dolomite are as follows: 

96.46 per cent CaC0 3 + MgCO 8 

2.21 per cent SiO 2 
10.23 per cent MgO 

1.33 per cent H 2 



CALCULATIONS FROM REPORTED PERCENTAGES 139 

The analysis of the lime shows no water and 1.37 per cent carbon dioxide. 
Calculate the percentages of CaO, MgO, and SiC>2 in the lime. 
Ans. CaO = 76.14, MgO = 18.52, SiO 2 = 4.00, CO 2 = 1.37. 

366. The oil in a sample of paint is extracted, and the residual pigment is 
found to be 66.66 per cent of the original weight. An analysis of the pigment 
gives 

Zinc oxide = 24.9 per cent 

Lithopone = 51.6 per cent 

Barium chromate = 23.5 per cent 

100.0 per cent 

Calculate the percentage composition of the original paint. 

Ans. Zinc oxide = 16.6, lithopone = 34.4, barium chromate = 15.7, 
oil = 33.3. 

366. The same sample of iron ore is analyzed by two chemists. Among 
other constituents, Chemist A reports: H 2 O = 1.62 per cent, Fe = 43.92 per 
cent. Chemist B reports: H 2 O = 0.96 per cent, Fe = 44.36 per cent. Cal- 
culate the percentage of iron in both cases on a dry sample. Analysis of a 
dry sample shows A to be correct. What are the error and percentage error 
in the constituent iron in B's analysis as reported? 

Ans. A = 44.64 per cent, B = 44.79 per cent. 0.15 per cent, 0.34 percent. 

367. A cargo of wet coal is properly sampled and the loss in weight at 
105C. is determined as 10.60 per cent. The dried sample is used for the 
analysis of other constituents, as follows: 

Volatile combustible matter = 21.60 per cent 

Coke = 60.04 per cent 

Ash = 18.36 per cent 

The air-dried coal (moisture content = 1.35 per cent) costs $8.80 a ton at 
the mine. What is its percentage of ash? Neglecting other factors except 
water content, calculate the value of the wet coal. 
Ans. 18.11 per cent. $7.98 a ton. 

368. The moisture content of a sample of A1 2 (SO4)3.18H 2 O is reduced from 
the theoretical to 7.36 per cent. Calculate the analysis of the partly dried 
material reporting percentage of Al 2 Os, SOs, and JH^O. The original salt 
costs $0.0262 per pound. Calculate the cost of the dried material, considering 
only the loss in water content. 

Ans. A1 2 O 3 = 27.64 per cent, SO 3 = 64.98 per cent, H 2 O = 7.36 per cent. 
$0.0473. 

369. In the paper industry "air-dry" paper pulp is considered as containing 
10 per cent of water. A sample of wet pulp weighs 737.1 grams and when 
heated to "bone dryness" weighs 373.6 grams. What is tne percentage of 
air-dry pulp in the original sample? 

Ans. 56.30 per cent. 



140 CALCULATIONS OF ANALYTICAL CHEMISTRY 

370. Ignition in air of MnOa converts it quantitatively into Mn 3 04. A 
sample of pyrolusite is of the following composition: MnO2 = 80.0 per cent; 
SiO 2 and other inert constituents = 15.0 per cent; H 2 O = 5.0 per cent. The 
sample is ignited in air to constant weight. Calculate the percentage of Mn 
in the ignited sample. 

Ans. 59.4 per cent. 

371. A salt mixture is found to contain 60.10 per cent UO 3 (essentially 
in the form of ammonium diuranate). It is also found that 10.00 per cent of 
the mixture is combined and uncombined volatile matter (essentially NH 3 
and H 2 O) and 29.90 per cent is nonvolatile inert matter. What is the per- 
centage of the clement uranium in the material after ignition if the volatile 
matter is all lost and the uranium is converted to the oxide U 3 O 8 ? 

Ans. 56.27 per cent. 



372. A sample of a mineral containing water as its only volatile constituent 
contains 26.40 per cent SiO 2 and 8.86 per cent water. What would be the 
percentage of SiO 2 in the material after heating sufficiently to drive off all the 
water, assuming that no chemical changes occur? What would be the percentage 
of SiO 2 if the ignited material still showed the presence of 1.10 per cent water? 

373. A powder consisting of a mixture of pure CuS0 4 .5H 2 O and silica con- 
tains 18.10 per cent Cu. What would be the percentage of combined sulfur in 
the material after the water of crystallization is all driven off by ignition? 

374. One pound of an ore lost 0.500 ounce of water by drying at 110C. to 
constant weight. The dried ore upon strong ignition with a flux lost 1.50 
per cent of its weight as moisture and was also found to contain 20.10 per 
cent SiO 2 . Find the percentage of water and of silica in the ore as received. 

376. A sample of crude copperas (FeSO 4 .7H 2 O) representing a large ship- 
ment was purchased at 1.25 cents per pound. An analysis for iron content 
gave 20.21 per cent Fe. The shipment was stored for a considerable period 
during which time water of crystallization was lost. To fix the price at which 
the copperas was to be sold, it was found that an increase of 0.023 cent per 
pound would be necessary, due entirely to the change in the percentage of 
iron. Calculate the percentage of iron in the sample after storage. Assume 
the increase to be due wholly to loss of water. 

376. A sample of dolomite analyzes as follows: 

SiO 2 = 0.31 per cent 
A1 2 O 3 = 0.07 per cent 
Fe 2 O 3 = 0.09 per cent 
MgO = 21.54 per cent 

CaO = 30.52 per cent 

CO 2 = 47.55 per cent 

The dolomite is ignited, and a 5.00-gram sample of the ignited material shows 
the presence of 0.80 per cent CO 2 . What weight of Mg 2 P 2 O 7 could be obtained 
from a 0.500-gram sample of the ignited material? 



CALCULATIONS FROM REPORTED PERCENTAGES 141 

377. A shipment of meat scrap is sold with the specification that it con- 
tains a minimum of 55.00 per cent protein and a maximum of 10.00 per cent 
fat when calculated to a dry basis. The analyst for the sender reports 53.20 
per cent protein, 9.59 per cent fat, and 3.60 per cent moisture. The material 
takes on moisture during shipment and the analyst for the receiver reports 
50.91 per cent protein, 9.31 per cent fat, and 7.50 per cent moisture. Do either 
or both of the analyses show conformity of the material to specifications? 
On the dry basis, what is the percentage variation between the two protein 
values and between the two fat values as reported by the analysts? 

378. A sample of limestone analyzes as follows: 

CaCO 3 = 86.98 per cent 

MgCO 3 = 3.18 per cent 

Fe 2 O 3 = 3. 10 per cent 

A1 2 O 3 = 0.87 per cent 

SiO 2 = 5.66 per cent 

H 2 O = 0.30 per cent 

100.09 per cent 

Analysis of the ignited material shows no moisture and only 1.30 per cent CO2. 
What is the percentage of Fe in the ignited material? 

379. A sample of pyrolusite analyzes as follows: MnO 2 = 69.80 per cent; 
SiO 2 and other inert constituents = 26.12 per cent; CO 2 = 1.96 per cent; 
H 2 O = 2.15 per cent. On ignition in air all the H 2 O and CO 2 are lost arid the 
MnO 2 is converted to Mn 3 O 4 . Calculate the percentage of Mn 3 O 4 in the ignited 
material. 

380. A carbonate rock analyzes as follows: CaO = 43.18 per cent; MgO = 
8.82 per cent; FeO = 3.10 per cent; Fe 2 O 3 = 1.90 per cent; SiO 2 = 7.30 per 
cent; CO 2 = 33.69 per cent; H 2 = 2.00 per cent. A portion of this rock is 
ignited and a sample of the ignited material on analysis shows 2.00 per cent 
CO2 and no water. It also shows that the ferrous iron has been completely 
oxidized. Calculate the percentage of total Fe 2 Os and of CaO in the ignited 
material. 

62. Calculation of Molecular Formulas from Chemical Analy- 
ses. Given a compound of unknown composition, a chemical 
analysis will determine the proportion in which the constituents 
of the compound exist. The results of such an analysis may then 
be used to calculate the empirical formula of the compound. 
Thus, the analysis of a certain salt gives the following results: 

Zinc = 47.96 per cent 

Chlorine = 52.04 per cent 

lOOOO per cent 

Dividing the percentage of each constituent by its atomic weight 
will give the number of gram-atoms of that constituent in 100 



142 CALCULATIONS OF ANALYTICAL CHEMISTRY 

grams of the compound. In 100 grams of the above salt there are 
present 47.96/65.38 = 0.7335 gram-atom of zinc and 52.04/35.46 = 
1.4674 gram-atoms of chlorine. These numbers are seen to be 
in the ratio of 1 to 2. The empirical formula of the salt is there- 
fore ZnCl'2, although, as far as the above analysis is concerned, the 
actual formula might be Zn 2 Cl4, ZnaCle, or any other whole mul- 
tiple of the empirical formula. Tn general, the determination of 
the molecular weight of a compound is necessary in order to deter- 
mine which multiple of the empirical formula will give the actual 
formula. The usual methods of establishing molecular weights 
by means of vapor density, freezing-point lowering, boiling-point 
raising, and other physicochemical phenomena should already be 
familiar to the student but the following will serve as a brief review. 

Equal volumes of gases under identical conditions of tempera- 
ture and pressure contain the same number of molecules (Avo- 
gadro). Therefore the molecular weights of gases are proportional 
to their densities. Since under standard conditions of tempera- 
ture and pressure (0C., 760 mm.) a gram-molecular weight of a 
gas (e.g., 32 grams of O 2 ; 28.016 grams of N 2 ) occupies 22.4 liters, 
an experimental method of determining the molecular weight of 
a gas is to measure its density under known conditions of tem- 
perature and pressure and calculate the weight of 22.4 liters of 
it under standard conditions (see Sec. 109). The molecular weight 
of a solid or liquid can also be determined in this way if the sub- 
stance can be converted to a gas without decomposition or change 
in degree of molecular association. 

A soluble substance lowers the freezing point and raises the 
boiling point of a definite weight of a solvent in proportion to 
the number of molecules or ions of solute present. In the case 
of a nonpolar (un-ionized) solute dissolved in water, one gram- 
molecular weight of the solute dissolved in 1,000 grams of water 
raises the boiling point of the water by 0.52C. (i.e., to 100.52C.) 
and lowers the freezing point of the water by 1.86 (i.e., to 
1.86C.). In general, for aqueous solutions of nonpolar solutes. 



Grams of solute . . 1,000 



Mol. wt. of solute grams of water 



X 0.52 = raising of 

boiling point 

X 1.86 = lowering of 

freezing point 



CALCULATIONS FROM REPORTED PERCENTAGES 143 

Ionized solutes change the boiling point or freezing point of a 
solvent to a greater degree owing to the greater number of par- 
ticles present. Thus, NaCl at ordinary concentrations depresses 
the freezing point of water and raises the boiling point by about 
twice as much as calculated from the above formula, owing to 
ionization into Na+ and Cl~ ions. Similarly, CaCl2 and Na 2 SO 4 
give an effect about three times as great as that of a nonpolar 
solute. 

Solutes dissolved in solvents other than water show analogous 
behavior in that the changes in freezing point and boiling point 
brought about by a mole of solute in 1,000 grams of solvent are 
fixed values (but of course different from those values given by 
water as the solvent). 

The determination of the molecular weight of a soluble sub- 
stance can therefore be made by preparing a solution of a known 
weight of it in a known weight of solvent (preferably water) and 
measuring the point at which the solution begins to freeze or boil. 

EXAMPLE 1. A certain organic compound is found by analysis 
to contain 40.00 per cent carbon, 6.71 per cent hydrogen, and the 
rest oxygen. When converted to a gas it has a density 2.81 times 
that of oxygen at the same temperature and pressure. What is 
the formula of the compound? 

SOLUTION: In 100 grams of substance there are 40.00 grams 
of C, 6.71 grams of H, and 53.29 grams of O. This corresponds to 

40.00 00 , . 

= 3.33 gram-atoms of C 



12.01 

6.71 
1.008 

53.29 



= 6.66 gram-atoms of H 
= 3.33 gram-atoms of 



16.00 

These are in the ratio of 1:2:1 

Empirical formula = CH 2 O 

Formula weight of CH 2 = 30 (approx.) 

Molecular weight of compound = 2.81 X 32 = 90 (approx.) 

Therefore, 

Formula of compound = CaHeOa. Ans. 



144 CALCULATIONS OF ANALYTICAL CHEMISTRY 

EXAMPLE II. Butandione is a yellow liquid containing 55.8 
per cent carbon, 7.00 per cent hydrogen, and 37.2 per cent oxygen. 
It is soluble in water, and the solution does not conduct electricity. 
A water solution containing 10.0 grams of the compound in 100 
grams of water freezes at 2.16C. What is the formula of 
butandione? 
SOLUTION: 

r r n 

Gram-atoms per 100 grams = r^ = 4.65 of C 

J Z.U 

-^-7.00 ofH 
- f?_ 2.32 of O 

These are in the approx. ratio of 2:3: 1 

Empirical formula = C^HaO 
Formula weight of C 2 H 3 O = 43 

10 ' X 1^X1.80 = 2.16 



Mol. wt. 100 
Solving, 

Mol. wt. = 86 
Therefore, 

Actual formula = C^Hf^. Ans. 

63. Calculation of Empirical Formula of a Mineral. The cal- 
culation of molecular formulas plays an important part in the 
analysis of natural minerals. A careful analysis furnishes a means 
of establishing the empirical formula of a mineral of high degree 
of purity, although its actual formula is usually impossible to 
determine by ordinary physicochemical methods since minerals 
cannot be vaporized or dissolved unchanged. The method of 
calculation is similar to that of the preceding examples, except 
that the basic constituents of a mineral are usually expressed in 
terms of their oxides. If the percentage of each constituent is 
divided by its molecular weight, the number of moles (gram- 
molecular weights) of that constituent in 100 grams of the mineral 
is obtained. From the ratios of the number of moles of the various 
constituents thus obtained, the formula of the mineral may be 
determined. It should be remembered, however, that analytical 



CALCULATIONS FROM REPORTED PERCENTAGES 145 

methods are subject to errors. It can hardly be expected, there- 
fore, that the number of moles of the various constituents as 
determined analytically will be exactly in the ratio of small whole 
numbers, although in the actual molecule (except in cases involv- 
ing isomorphism, discussed below) the molar ratios are small whole 
numbers. In a few cases, some judgment must be exercised in 
order to determine from the analysis the true molar ratios of the 
constituents in the molecule. A slide rule will be found to be 
almost indispensable for this purpose, since, with two settings of 
the rule, all possible ratios are visible. 

EXAMPLE. The analysis of a certain mineral gives the follow- 
ing results : 

A1 2 O 3 - 38.07 per cent 

K 2 O = 17.70 per cent 

CaO = 10.46 per cent 

SiO 2 = 3370 per cent 
99.93 per cent 

What is the empirical formula of the mineral? 

SOLUTION: In 100 grams of the mineral there are present 

38.07 38.07 



A1 2 O 3 102.0 
17.70 17.70 
K 2 O 94.20 
10.46 10.46 
CaO ~ 56.07 

33.70 = 33.70 
"SiO 2 " 60.3 



= 0.3733 mole of A1 2 3 
= 0.1879 mole of K 2 O 
-0.1865 mole of CaO 
= 0.559 mole of SiO 2 



It is seen that the moles of these constituents are near enough 
in the ratio of 2:1:1:3 to be within the limits of experimental 
error. The molecule is therefore made up of 2A1 2 O 3 .K 2 0. Ca0.3SiO 2 
and may be written K 2 CaAl 4 Si 3 Oi4. 

64. Calculation of Formulas of Minerals Exhibiting Isomorphic 
Replacement. Complications arise in the calculation of formulas 
in the cases of minerals exhibiting isomorphic replacement, i.e., 
the partial replacement of one constituent by one or more other 
constituents having the same general properties. It therefore 
happens that, owing to different degrees of replacement, samples 



146 CALCULATIONS OF ANALYTICAL CHEMISTRY 

of the same kind of mineral obtained from different localities often 
give on analysis numerical results which apparently bear little 
resemblance to one another. 

As a general rule, a constituent may be replaced only by another 
of the same type and valence. Thus, Fe 2 3 is often partially or 
wholly replaced by A1 2 O 3 , and vice versa. CaO may be replaced 
by MgO, MnO, FeO, etc. Exceptions are sometimes met with, 
but, for purposes of calculation, this assumption may be safely 
made. Since the isomorphic replacement occurs in no definite 
proportion, it follows that the molar amounts of the constituents 
in such minerals do not necessarily bear any simple relation to 
one another. On the other hand, if constituent B partially re- 
places constituent A, since the valences are the same, the sum of 
the molar amounts of A and B would be the same as the molar 
amount of A if it had not been replaced. Consequently, when 
the molar quantities of the constituents of a mineral in them- 
selves bear no simple ratio to one another, the quantities of con- 
stituents of the same type should be combined in an effort to 
obtain sums that do exist in ratios of simple whole numbers. 

EXAMPLE. A certain mineral gives the following analysis: 



A1 2 3 = 20.65 per cent 
Fe 2 O 3 = 7.03 per cent 
CaO = 27.65 per cent 
Si0 2 = 44.55 per cent 
99J88 per cent 



What is the empirical formula? 

SOLUTION: The number of moles of each constituent in 100 
grams of the mineral is found to be 

= 0.2025 mole of A1 2 O 3 ( 

,= 0.2465 mole 

4rr = 0.0440 mole of Fe 2 8 
reaUa 

27.65 



CaO 

44.55 
Si0 2 



= 0.4932 mole of CaO 
= 0.7389 mole of Si0 2 



CALCULATIONS FROM REPORTED PERCENTAGES 147 

Only when the molar quantities of the first two constituents 
are combined are all the above numerical results found to be in 
simple ratio to one another, these being approximately as 1:2:3. 
This shows isomorphic replacement between Fe20 3 and AhOa, 
and the formula of the mineral may therefore be written 

(Al,Fe) 2 O 3 .2CaO.3Si0 2 
or 

Ca 2 (Al,Fe) 2 Si 3 Oii. Ans. 

Problems 

381. From the following percentage composition of ethylamine, calculate 
its empirical formula: C = 53.27 per cent, H = 15.65 per cent, N = 31.08 
per cent. 

Ans. C 2 H 7 N. 

382. Calculate the empirical formula of the compound having the following 
composition: Ca = 23.53 per cent, H = 2.37 per cent, P = 36.49 per cent, 
O = 37.61 per cent. 

Ans. Ca(H 2 P0 2 ) 2 . 

383. Calculate the empirical formula of an organic compound having the 
following composition: C = 68.83 per cent, H == 4.96 per cent, O = 26.21 per 
cent. 

Ans. CylleOjN 

384. Show that the following analysis of diethylhydrazine agrees with 
the formula (C 2 H 6 ) 2 :N.NH 2 : 



Carbon = 54.55 per cent 

Hydrogen = 13.74 per cent 

Nitrogen = 31.80 per cent 

100.09 per cent 



386. Calculate the molecular formula of a compound that has a molecular 
weight of approximately 90 and has the following composition: C = 26.67 per 
cent, H = 2.24 per cent, O = 71.09 per cent. 

Ans. H 2 C 2 O 4 . 

386. A certain compound of carbon and oxygen has an approximate 
molecular weight of 290 and by analysis is found to contain 50 per cent by 
weight of each constituent. What is the molecular formula of the compound? 

Ans. Ci 2 Og. 

387. What is the molecular weight of a substance, 0.0850 gram of which 
dissolved in 10.0 grams of water gives a solution freezing at 0.465C. and 
not conducting electricity? What is the molecular formula of the substance 
if it contains 5.94 per cent hydrogen and 94.06 per cent oxygen? 

Ans. 34.0. H 2 2 . 



148 CALCULATIONS OF ANALYTICAL CHEMISTRY 

388. A certain organic compound contains 48.64 per cent carbon, 43.19 
per cent oxygen, and the rest hydrogen. A solution of 7.408 grams of the solid 
in 100 grams of water boils at 100.52C. and does not conduct electricity. 
What is the molecular formula of the compound? 

Ans. C,H 6 O 2 . 

389. A certain gaseous compound is found by analysis to consist of 87.44 
per cent nitrogen and 12.56 per cent hydrogen. If 500 ml. of the compound 
has almost exactly the same weight as that of 500 ml. of oxygen at the same 
temperature and pressure, what is the molecular formula of the compound? 

Ans. N 2 II 4 . 

390. A certain organic compound contains almost exactly 60 per cent 
carbon, 5 per cent hydrogen, and 35 per cent nitrogen. A solution of 20.0 
grams of the compound in 300 grams of water does not conduct electricity 
and freezes at 1.55C. What is the molecular formula of the com- 
pound? 

Ans. 



391. A certain sugar is a compound of carbon, hydrogen, and oxygen. 
Combustion in oxygen of a sample weighing 1.200 grams yields 1.759 grams of 
C02 and 0.720 gram of H 2 O. A solution of 8.10 grams of the substance in 150 
grams of water boils at 100.156C. What is the molecular formula of the sugar? 
At what temperature should the above solution freeze? 

Ans. C 6 H 12 O 6 . -0.558C. 

392. At a certain temperature and pressure 250 ml. of a certain gas con- 
sisting of 90.28 per cent silicon and 9.72 per cent hydrogen has a weight equal 
to that of 555 ml. of nitrogen at the same temperature and pressure. What 
is the molecular formula of the gas? 

Ans. Si 2 H 6 . 

393. When 0.500 gram of a certain hydrocarbon is completely burned in 
oxygen, 0.281 gram of H 2 O and 1.717 grams of CO 2 are formed. When a 
certain weight of this compound is vaporized, it is found to have a volume 
almost exactly one-quarter that of the same weight of oxygen under the same 
conditions of temperature and pressure. Calculate the molecular formula of 
the compound. 

Ans. dolis. 

394. An analysis of a mineral gave the following results: H 2 = 4.35 per 
cent, CaO = 27.15 per cent, A1 2 O 3 = 24.85 per cent, SiO 2 = 43.74 per cent. 
Calculate the empirical formula of the mineral. 

Ans. H 2 Ca 2 Al 2 Si 3 Oi 2 . 

395. Dana gives the composition of vivianite as follows: P 2 Os 28.3 per 
cent, FeO = 43.0 per cent, H 2 = 28.7 per cent. Show that this conforms to 
the formula 



CALCULATIONS FROM REPORTED PERCENTAGES 149 

396. The percentage composition of a certain silicate is as follows: K 2 O = 
21.53 per cent, A1 2 O 3 = 23.35 per cent, Si0 2 = 55.12 per cent. Calculate the 
empirical formula of the silicate. 

Ans. KAlSi 2 O 6 . 

397. A certain compound contains only the following constituents: CaO, 
Na 2 O, SO 8 . The percentages of these constituents are in the respective 
approximate ratios of 9:10:26. What is the empirical formula of the com- 
pound? 

Ans. Na 2 CaS 2 8 . 

398. What is the empirical formula of a simple basic cupric carbonate 
which, according to Rogers, contains 57.4 per cent Cu and 8.1 per cent H 2 O? 

Ans. Cu 2 (OH) 2 CO 3 . 

399. The composition of bismutite is given by Ramm as follows: CO 2 = 
6.38 per cent, Bi 2 O 3 = 89.75 per cent, H 2 O = 3.87 per cent. Calculate the 
empirical formula. 

Ans. 2Bi 8 C 3 Oi 8 .9H 2 O. 

400. What is the empirical formula of a silicate which gives the following 
analysis: 

CaO = 24.72 per cent 

MgO = 11. 93 per cent 

FeO = 10.39 per cent 

SiO 2 = 53.09 per cent 

100.13 per cent 

Ans. Ca(Mg,Fe)(SiO 3 ) 2 . 

401. A silicate gives the following analysis. Tf two-thirds of the water 
exists as water of crystallization, what is the empirical formula? 

H 2 O = 17.22 per cent 

CaO = 8.22 per cent 

Na*O = 0.76 per cent 

A1 2 O 3 = 16.25 per cent 

SiO 2 = 57.48 per cent 

99.93 per cent 

Ans. H 4 (Ca,Na 2 )Al 2 (Si0 3 ) 6 .4H 2 0. 

402. Calculate the empirical formula of a mineral that analyzes as follows: 

MnO = 46.36 per cent 

CaO = 6.91 per cent 

Si0 2 = 46.78 per cent 

100.05 per cent 

Ans. (Mn,Ca)SiO 8 . 

403. The analysis of samples of microcline and of albite are given below. 
Show that these minerals are of the same type. Give the general empirical 
formula. Assume the percentages of silica and alumina to be the most reliable. 



150 CALCULATIONS OF ANALYTICAL CHEMISTRY 

MICROCLINE ALBITB 

Na 2 O = 1.61 per cent Na2O = 11.11 per cent 

K 2 O = 13.56 per cent K 2 O = 0.51 per cent 

A1 2 O 3 = 19.60 per cent CaO = 0.38 per cent 

SiO 2 = 64.79 per cent A1 2 O 3 = 19.29 per cent 

99.56 per cent SiO 2 = 68.81 per cent 

100.10 per cent 
Ans. (K,Na)AlSi 3 O 8 . 

404. Calculate the empirical formula of axinite from the following analysis: 

H 2 O = 1.58 per cent 

CaO = 19.63 per cent 

FeO = 9.54 per cent 

MnO = 3.01 per cent 

A1 2 O 3 = 17.92 per cent 

B 2 O 3 = 6.12 per cent 

SiO 2 = 42.23 per cent 
100.03 per cent 

Ans. HCa 2 (Fe,Mn)Al 2 B(SiO 4 )4. 

406. A sample of the mineral biotite gave the following analysis: 

H 2 O = 1.10 per cent 

FeO = 9.60 per cent 
A1 2 O 3 = 22.35 per cent 

K 2 O = 14.84 per cent 
MgO = 12.42 per cent 

SiO 2 = 39.66 per cent 
99.97 per cent 

What is the empirical formula of the mineral? 
Ans. (H,K) 2 (Mg,Fe) 2 Al 2 Si 3 12 . 



406. What is the empirical formula of a compound of the following composi- 
tion: K = 38.68 per cent, H = 0.50 per cent, As = 37.08 per cent, O = 23.74 
per cent? 

407. Analysis of anorganic compound gave the folio wing results : C = 60.86 
per cent, H = 4.38 per cent, O = 34.76 per cent. Calculate the empirical 
formula of the compound. 

408. An organic acid is found to have a molecular weight of approximately 
160 and to give the following analysis: carbon = 57.82 per cent, hydrogen = 
3.64 per cent, oxygen (by difference) = 38.54 per cent. Calculate the 
molecular formula of the acid. 

409. Calculate the empirical formula of the compound of the following 
composition: Sb = 49.55 per cent, O = 6.60 per cent, Cl = 43.85 per cent. 

410. Metaformaldehyde contains 40.00 per cent carbon, 6.67 per cent hydro- 
gen, and 53.33 per cent oxygen. A solution of 10.01 grams of the compound 



CALCULATIONS FROM REPORTED PERCENTAGES 151 

in 250 grams of water freezes at 0.827C. and does not conduct electricity. 
What is the molecular formula of metaformaldehyde? 

411. A certain derivative of benzene contains only carbon, hydrogen, and 
nitrogen, and is not ionized in aqueous solution. Analysis of the compound 
shows 58.51 per cent carbon and 7.37 per cent hydrogen. A solution of 30.0 
grains of the compound in 150 grams of water freezes at 3.02C. What is 
the molecular formula of the compound? 

412. Wliat is the molecular weight of a substance 4.50 grams of which, 
when dissolved in 50.0 grams of water, gives a solution freezing at 0.93C. 
but not conducting electricity? What is its molecular formula if it contains 
40.00 per cent carbon, 6.67 per cent hydrogen, and 53.33 per cent oxygen? 
At what temperature would the solution boil? 

413. A certain hydrocarbon contains 79.89 per cent carbon and has a 
density 1.07 times that of N 2 at the same temperature and pressure. What is 
the molecular formula of the hydrocarbon? 

414. A sample of a certain compound of carbon, hydrogen, and oxygen 
weighing 2.000 grains yields on combustion in oxygen 2.837 grams of CO 2 and 
1.742 grams of H 2 O. A solution of 2.150 grams of the compound in 50.0 grams 
of water is nonconducting and freezes at 1.288C. What is the molecular 
formula of the compound? 

416. Under standard conditions a liter of a certain gaseous compound of 
boron and hydrogen weighs 2.38 grains. When 1.00 gram of this compound is 
heated, it is completely decomposed into boron and hydrogen, and the latter 
has a volume of 2.10 liters under standard conditions. Calculate the molecular 
formula of the boron hydride. 

416. A certain gas is composed of 46.16 per cent carbon and the remainder 
nitrogen. Its density is 1.80 times that of air at the same temperature and 
pressure. What is the molecular formula of the gas? (Calculate the apparent 
molecular weight of air by considering it four-fifths nitrogen and one-fifth 
oxygen.) 

417. Calculate the formula of a compound of carbon, hydrogen, nitrogen, 
and oxygen from the following data. Approximate mol. wt. 140. Decom- 
position of a 0.2-gram sample gives 32.45 ml. of nitrogen when measured 
dry under standard conditions. The same weight of sample on combustion 
in oxygen yields 0.3824 gram of CO 2 and 0.0783 gram of I1 2 O. 

418. What is the empirical formula of a mineral containing 3.37 per cent 
water, 19.10 per cent aluminum oxide, 21.00 per cent calcium oxide, and 
56.53 per cent silica? 

419. Zircon is a pure silicate of zirconium containing 33.0 per cent of silica. 
What is its empirical formula? 

420. A tungstate has the following composition: WO 3 = 76.5 per cent, 
FeO = 9.5 per cent, MnO = 14.0 per cent. Calculate the empirical formula. 

421. Calamine is a basic zinc silicate of the following composition: ZnO = 
67.5 per cent, H 2 O = 7.5 per cent, SiO 2 = 25.0 per cent. Calculate its em- 
pirical formula. 



152 CALCULATIONS OF ANALYTICAL CHEMISTRY 

422. A silicate of the composition given below is found to have 85 per cent 
of its water in the form of water of crystallization. What is the empirical 
formula? 

H 2 = 7.7 per cent 
KO = 28.1 per cent 
CaO = 20.4 per cent 
SiO 2 = 43.8 per cent 
100.0 per cent 

423. A sample of a certain hydrogen-potassium-magnesium-aluminum 
silicate weighing 1.2000 grams yields the following products: 0.0516 gram 
of water, 0.4000 gram of KC10 4 , 0.9550 gram of Mg 2 P 2 O 7 , and 0.1461 gram 
of alumina. What is the empirical formula of the mineral? 

424. From the following data obtained from the analysis of a feldspar, 
calculate the percentage composition of the sample and determine the em- 
pirical formula of the mineral, omitting the calcium from the formula and 
assuming the percentages of silica and alumina to be the most reliable. 

Sample taken = 1 .2000 grams 

Silica obtained = 0.7751 gram 

Alumina obtained = 0.2255 gram 

Calcium oxide obtained = 0.0060 gram 
KC1 + NaCl obtained = 0.3193 gram 
K 2 PtCl 6 obtained = 0.7240 grain 



PART III 
VOLUMETRIC ANALYSIS 

CHAPTER XI 
CALIBRATION OF MEASURING INSTRUMENTS 

65. Measuring Instruments in Volumetric Analysis. The prin- 
ciple of volumetric analysis differs from that of gravimetric analysis 
in that, instead of isolating and weighing a product of a reaction 
directly or indirectly involving the desired substance, the volume 
of a reagent required to bring about a direct or indirect reaction 
with that substance is measured. From the volume of the reagent 
and its concentration, the weight of the substance is calculated. 

Since volumetric analysis makes use of exact volume relation- 
ships, it is essential first to adopt a definite standard for a unit 
volume and then to calibrate all measuring instruments to con- 
form to this standard. The measuring instruments most often 
used are burets, pipets, and measuring flasks, and the experi- 
mental methods of calibrating them may be found in any standard 
reference book on quantitative analysis. 

66. Calculation of True Volume. A liter is the volume occupied 
by 1 kilogram of water at the temperature of its maximum den- 
sity (approximately 4C.). A milliliter (ml.} is 1/1,000 liter. A 
cubic centimeter (cc.) is the volume occupied by a cube 1 cm. on 
a side. One liter contains 1000.027 cc. In calibrating a vessel, 
since the cubical content of the vessel holding the water to be 
weighed varies with the temperature, it is evident that the tem- 
perature of the container must be included in the specifications. 
Instead of taking the corresponding temperature of 4C., the tem- 
perature of 20C. has been accepted as the normal temperature by 
the Bureau of Standards at Washington. 

To contain a true liter then, a flask must be so marked that at 
20C. its capacity will be equal to the volume of water which 

153 



154 CALCULATIONS OF ANALYTICAL CHEMISTRY 

at 4C. weighs 1 kilogram in vacuo. From the density of water 
at different temperatures (Table IV, Appendix), the coefficient 
of cubical expansion of glass (0.000026), and the relationship 
existing between the weight of a substance in air and the weight 
in vacuo (Sec. 49), it is possible to calculate the amount of water 
to be weighed into a container in order that it shall occupy a true 
liter at any given temperature. 

EXAMPLE. How much water at 25C. should be weighed in 
air with brass weights so that when placed in a flask at the same 
temperature and under normal barometric pressure it will occupy 
1 true liter at 20C.? 

SOLUTION: Density of water at 25C. = 0.99707 (Table TV). 

At 4C. and in vacuo, 1,000 grams of water will occupy 1 true liter. 

At 25C. and in vacuo, 1,000 X 0.99707 grams of water will 
occupy 1 true liter. 

At 25C. and in air, the weight of water is found by substituting 

in the formula 

(W W\ 
?-?> 

and solving for W (see Sec. 49). 
Thus, 



Since the term to the right of the plus sign is required to only 
two significant figures, it is sufficiently accurate to write 

997.07 = W 

whence 

W = 996.02 grams 

Theoretically, to contain a true liter, the flask must be at 20C. 
and yet contain this weight of water at 25C. Actually, the tem- 
perature of the flask is also 25C. It has therefore expanded, the 
cubical content is greater, and the true-liter volume is also greater. 
The coefficient of cubical expansion of glass is 0.000026, and the 
increase in volume from 20 to 25C. is 1,000 X 0.000026(25 - 20) = 
0.13 ml. This volume is represented by 0.13 X 0.99707 = 0.13 
gram of water. The required weight of water is therefore 

996.02 + 0.13 = 996.15 grams. Ans. 



CALIBRATION OF MEASURING INSTRUMENTS 155 

A general formula may now be written for calculating the weight 
of water required for a true liter. 

W = 1>00 X d + [1,000 X d X c(t - 20)] 

1 4- a a 
1 T" ~j J7 

d d 

where W = grams of water required for 1 true liter 
t = temperature of water and flask 
d = density of water at t 

a = weight of 1 ml. of air under given conditions 
d' = density of balance weights 
c = coefficient of cubical expansion of the container 

(The values of these last three terms are usually 0.0012, 8.0, and 
0.000020, respectively.) 

The correction for the expansion or contraction of the container 
is in each case small compared with the quantity to which it is 
added. Consequently, only an approximate value containing two 
or three significant figures need be used. Indeed, in the case of 
instruments of 50-ml. content or less and for small differences in 
temperature, this correction may ordinarily be neglected. 

By using the third column of Table IV, calculations like the 
above can be simplified. This column gives the weight of 1 ml. 
of water at a given temperature when the weighing is made in 
air against brass weights and the water is in a glass container. 
In other words, corrections for expansion of glass and for con- 
version to vacuo are incorporated in the values given. It is seen, 
for example, that the answer to the above problem is found di- 
rectly by multiplying by 1,000 the weight of 1 ml. of water at 
25C. under the conditions specified. 

0.99615 X 1,000 = 996.15 grams. Ans. 

Problems 

426. Calculate accurately the amount of water which should be weighed 
into a tared flask at 18C. and 770 mm. pressure against brass weights in 
order that the flask may be marked to contain exactly 250 true milliliters. 

Ans. 249.37 grams. 

426. A flask which has been marked to contain 1 true liter is filled with 
water at 15C. to the mark, and the temperature of the water is allowed to 



156 CALCULATIONS OF ANALYTICAL CHEMISTRY 

rise to 25C. How many millimeters above the mark does the water now 
stand (inner diameter of the neck of the flask = 15.0 mm.)? 
Ans. 10.2 mm. 

427. In calibrating a flask to contain 3^ true liter, if the water is weighed 
at 26C. against brass weights, what percentage error would be introduced 
in the weight of water necessary if the expansion of glass were neglected? 
What weight of water should be taken? 

Ans. 0.016 per cent. 498.07 grams. 

428. What is the true volume of a flask that contains 746.24 grams of 
water at 30C. when weighed in air against brass weights? 

Ans. 750.06 ml. 

429. A flask is accurately marked to contain 1 true liter, and the inner 
diameter of the neck of the flask is 16.0 mm. If 996.00 grams of water are 
weighed out in air against brass weights at 20C. and placed in the flask at this 
temperature, how far above or below the true-liter mark does the meniscus 
of the water lie? 

Ans. 5.9 mm. below. 



430. To calibrate a flask to contain a true liter at 20C., how much water 
at 31 C. and 760 mm. pressure must be weighed into the flask in air against 
brass weights? 

431. In calibrating a flask to contain J^ true liter using water at 30C., 
what percentage error would be introduced by neglecting the expansion of 
glass in the calculation of the weight of water required? 

432. What is the true volume of a flask that contains 398.70 grams of water 
at 26C. when weighed in air against brass weights? 

433. A pipet is marked to contain 100 ml. according to the true-liter 
standard. It is filled with water at 12C. How much would the water weigh 
in air under normal barometric pressure against gold weights? 

434. The inner diameter of the neck of a flask is 14 mm., and the flask 
contains 498.00 grams of water at 28C. weighed in air against brass weights. 
How far above or below the meniscus of the water should a mark be placed 
in order to represent a volume of 500 ml. according to the true-liter 
standard? 

436. A 50-ml. buret is calibrated by weighing in ah 1 (against brass weights) 
the water delivered between 10-ml. intervals in the graduations. Calculate 
from the following data the true volume of solution that the buret will deliver 
between each 10-ml. interval, calculate the true total volume delivered be- 
tween the 0- and the 50-ml. mark, and make a graph showing the correction 
that must be applied to the buret reading to obtain the true volume in any 
titration where the buret is initially filled to the zero mark (temperature of the 
water = 25C.). 



CALIBRATION OF MEASURING INSTRUMENTS 157 



IKTEKVALS, Mr,. 

0.03-10.07 10.04 

10.07-19.93 9.84 

19.93-29.97 10.07 

29.97-40.03 10.13 

40.03-49.96 10.05 

If this buret is used in a titration and the initial and final readings are 0.11 
and 46.38, respectively, what volume of solution has actually been delivered? 



CHAPTER XII 

NEUTRALIZATION METHODS 
(ACIDIMETRY AND ALKALIMETRY) 

67. Divisions of Volumetric Analysis. It is customary to di- 
vide the reactions of volumetric analysis into four groups, viz., 

1. Acidimetry and alkalimetry 

2. Oxidation and reduction ("redox") methods 

3. Precipitation methods 

4. Complex formation methods 

In Sees. 24 and 25 the principles underlying the use of equiva- 
lents, milliequivalents, and normal solutions were taken up in a 
general way. In this and succeeding chapters these principles are 
reviewed, developed, and applied to the above four types of volu- 
metric analysis. 

68. Equivalent Weights Applied to Neutralization Methods. 
The fundamental reaction of acidimetry and alkalimetry is as 

follows : 

H+ + Oil- - H 2 O 

i.e., the neutralization of an acid by a base, or the neutralization 
of a base by an acid. 

The gram-equivalent weight of a substance acting as an acid is 
that weight of it which is equivalent in total neutralizing power 
to one gram-atom (1.008 grams) of hydrogen as hydrogen ion. 
The gram-equivalent weight of a substance acting as a base is 
that weight of it which will neutralize one gram-atom of hydrogen 
ion. 

A normal solution of an acid or base contains one gram-equiva- 
lent weight of the acid or base in one liter of solution, or one 
gram-milliequivalent weight in one milliliter of solution (see also 
Sec. 24). 

When hydrochloric acid reacts as an acid, the gram-molecular 
weight (36.47 grams) of hydrogen chloride furnishes for the neu- 
tralization of any base one gram-atom (1.008 grams) of reacting 

158 



NEUTRALIZATION METHODS 159 

hydrogen. According to the definition, the value 36.47 grams con- 
stitutes the gram-equivalent weight of hydrogen chloride, and a 
liter of solution containing this amount is a normal solution of the 
acid. In this case, the normal solution and the molar solution 
are identical. On the other hand, the amount of hydrogen sulfate 
required to furnish in reaction one gram-atomic weight of hydrogen 
is only one-half the gram-molecular weight, or H 2 S0 4 /2 = 49.04 
grams, and a normal solution of sulfuric acid would contain 49.04 
grams of hydrogen sulfate per liter of solution. A molar solution 
of sulfuric acid is therefore 2 normal and contains 2 gram-equiva- 
lent weights per liter, or 2 gram-milliequivalent weights per milli- 
liter. 

Acetic acid, HC2H 3 2 , contains 4 hydrogen atoms in its mole- 
cule; but, when the compound acts as an acid, only one of these 
hydrogens is involved in active reaction, thus: 

HC 2 H 3 O 2 + OH- -> C 2 H 8 2 - + H 2 O 

Consequently, H0 2 II 3 2 /1 = 60.05 grams of acetic acid consti- 
tute the gram-equivalent weight, and the normal solution con- 
tains this weight of acid in a liter, or 0.06005 gram of acetic acid 
per milliliter. 

Sodium hydroxide is neutralized as follows : 

(Na+)OH- + H+ -> H 2 (+ Na+) 

NaOH/1, or 40.00 grams, of sodium hydroxide constitutes one 
gram-equivalent weight of the alkali, not because the molecule 
contains one atom of hydrogen, but because it involves in reaction 
one atomic weight of hydrogen ion, as shown in the above equa- 
tion. Therefore, a weight of 40.00 grams of sodium hydroxide in 
a liter of solution represents the normal solution. When calcium 
oxide is used as a base, each gram-molecule reacts with two gram- 
atoms of hydrogen, thus: 

CaO + 211+ -* Ca+ + + H 2 

or CaO/2 = 28.04 grams of calcium oxide are needed to involve 
in reaction one gram-atom of hydrogen. Therefore, 28.04 grams 
of calcium oxide constitute the gram-equivalent weight in this case, 
although calcium oxide in itself contains no hydrogen whatever. 

Total neutralizing power should not be confused with degree of 
ionization. Equal volumes of normal solutions of hydrochloric 



160 CALCULATIONS OF ANALYTICAL CHEMISTRY 

acid and acetic acid have the same total neutralizing power, but 
the acids have very different degrees of ionization. In other words, 
equivalent weight is based on neutralizing power and not on rela- 
tive " strength" or degree of ionization. 

Problems 

436. What is the equivalent weight of zinc oxide as a base? Of KHSO 
as an acid? 

An*. 40.69. 136.2 

437. What is the milliequivalent weight in grams of each of the following 
acids or bases, assuming complete neutralization in each case: (a) NaaCOj, 
(6) K 2 O, (c) NH 4 OH, (d) HBr, (e) H 2 SO 3 , (/) H,PO 4 ? 

Ans. (a) 0.05300 gram, (6) 0.04710 gram, (c) 0.03505 gram, (d) 0.08093 
gram, (e) 0.04103 gram, (/) 0.03267 gram. 

438. What is the equivalent weight of the following acids or bases, assuming 
complete neutralization in each case: (a) N 2 O 5 , (6) (NH 4 ) 2 O, (c) Ba(OII)2, 
(d) N1I 3 , (e) S0 2 ? 

Ans. (a) 54.01, (6) 26.05, (c) 85.69, (d) 17.03, () 32.03. 

439. What is the gram-milliequivalent weight of KsCOs in the reaction 
K 2 CO 3 + HC1 -> KHCOs + KC1? What is the gram-equivalent weight of 
H 3 PO 4 in the reaction H 8 PO 4 + 2NaOH -> Na 2 HPO 4 + 2H 2 O? 

Ans. 0.1382 gram. 49.00 grams. 

440. How many grams of oxalic acid, H 2 C 2 O 4 .2H 2 O, are required to make 
(d) a liter of molar solution, (b) a liter of normal solution, (c) 400 ml. of 
half-normal solution? 

A?is. (a) 126.1 grams, (b) 63.00 grams, (c) 12.61 grams. 

441. Formic acid (HCHO 2 ) is a monobasic acid that is 3.2 per cent ionized 
in 0.20 normal solution. What weight of the pure acid should be dissolved in 
250.0 ml. in order to prepare a 0.2000 normal solution? 

Ans. 2.301 grams. 

442. What weight of CaO is necessary to prepare the following: (a) 500 ml. 
of a one-hundredth molar solution of Ca(OH) 2 , (6) 30.63 ml. of N/100 Ca(OH) 2 ? 

Ans. (a) 0.2804 gram, (b) 0.008588 gram. 

443. What is the normality of a sulfurous acid solution containing 6.32 
grams of S0 2 per liter? Of an ammonium hydroxide solution containing 
17.5 grams of NH 3 in 480 ml. of solution? 

Ans. 0.197 N. 2.14 N. 

444. A solution of hydrochloric acid has a specific gravity of 1.200 and 
contains 39.11 per cent HC1 by weight. Calculate (a) the molar concentra- 



NEUTRALIZATION METHODS 161 

tion of the solution, (b) the normality of the solution, (c) the number of gram- 
equivalent weights of HC1 in every 750 ml. of solution. 
Ans. (a) 12.87 molar, (6) 12.87 N, (c) 9.65. 

446. A solution of sulfuric acid has a specific gravity of 1.100 and contains 
15.71 per cent H2SO4 by weight. What is the normality of the solution? 

Ans. 3.524 N. 

446. Assuming complete neutralization in each case, what are the equiv- 
alent weights of the following substances when acting as acids and bases? 
(a) LiOH, (b) H 2 SO 4 , (r) Fe 2 O 3 , (</) HC 2 H 3 O 2 , (e) cream of tartar (KHC 4 H 4 O 6 ). 

447. How many grams of II 2 SO 4 does a liter of 0.1000 N sulfuric acid 
solution contain? 

448. How many milliliters of HC1 (sp. gr. 1.200, containing 39.11 per cent 
HC1 by weight) are required to make a liter of N/10 solution by dilution with 
water? 

449. How many grams of hydrated oxalic acid (H 2 C 2 O 4 .2H 2 0) must be 
dissolved and diluted to exactly one liter to make a 0.1230 N solution for use 
as an acid? 

460. Chloracetic acid, CH 2 C1.COOH (mol. wt. = 94.50), is a monobasic 
acid with an ionization constant of 1.6 X 10~ 3 . How many grams of the acid 
should be dissolved in 300.0 ml. of solution in order to prepare a half-normal 
solution? 

451. How many grams of pure potassium tetroxalate (KHC 2 O 4 .H 2 C 2 O 4 .2H 2 O) 
must be dissolved in water and diluted to exactly 780 ml. to make a 0.05100 N 
solution for use as an acid? 

452. How many milliliters of sulfuric acid (sp. gr. 1.200, containing 27.32 
per cent H 2 SO 4 by weight) are required to make one liter of 0.4980 N solution 
by dilution with water? 

453. What is the normality of a sulfuric acid solution that has a specific 
gravity of 1.839 and contains 95.0 per cent H 2 SO 4 by weight? 

454. If 75.0 milliliters of hydrochloric acid (sp. gr. 1.100, containing 20.01 
per cent hydrochloric acid by weight) have been diluted to 900 ml., what is the 
normality of the acid? 

455. How could a solution of HC1 be prepared of such normality that each 
milliliter would represent 0.01000 gram of NaNO 2 when the latter is deter- 
mined by reducing to an ammonium salt, distilling the NH 3 with excess caustic 
alkali, and titrating the NH 3 with the standard HC1? 

456. A 0.2000 N solution of barium hydroxide is to be prepared from pure 
Ba(OH) 2 .8H 2 O crystals that have lost part of their water of crystallization. 
How may the solution be made if no standardized reagents are available? 
State specifically the treatment given and the weight and volume used. 

69. Normality of a Solution Made by Mixing Similar Com- 
ponents. When several similar components are mixed and dis- 



162 CALCULATIONS OF ANALYTICAL CHEMISTRY 

solved in water, the normality of the resulting solution is deter- 
mined by calculating the total number of equivalent weights 
present in a liter of solution. 

EXAMPLE. If 3.00 grams of solid KOH and 5.00 grams of solid 
NaOH are mixed, dissolved in water, and the solution made up 
to 1,500 ml., what is the normality of the solution as a base? 
SOLUTION: The number of equivalent weights of KOH in 1,500 
i . 3.00 3.00 T , r , ,, . 3.00 1,000 
mL 1S KOH = 5630" In l hter thcre 1S 500 X 



equivalent weight of KOH. In a liter of the solution there is 
also ^^\ x r?S = - 0833 equivalent weight of NaOH. A total 



of 0.0356 + 0.0833 = 0.1189 equivalent weight of base in a liter 
makes the normality of the solution as a base 0.1189 y N. Ans. 

Problems 

467. What is the normality of an alkali solution made by dissolving 6.73 
grams of NaOH (99.5 per cent NaOH, 0.5 per cent H 2 O) arid 9.42 grams of 
pure Ba(OII)2.8H 2 O in water and diluting to 850 ml.? 

Ans. 0.267 N. 

468. If 50.00 ml. of sulfuric acid (sp. gr. 1.420, containing 52.15 per cent 
of H 2 SO 4 by weight) and 50.00 ml. of sulfuric acid (sp. gr. 1.840, containing 
95.60 per cent H 2 SO4 by weight) are mixed and diluted to 1,500 ml., what is 
the normality of the solution as an acid? 

Ans. 1.699 N. 

459. If 50.00 ml. of a solution containing 5.000 grams of NaOH are added to 
50.00 ml. of a solution containing 5.000 grains of KOH, what is the normality 
of the mixture before and after dilution to 116.3 ml.? 

Arts. 2.141 N, 1.841 N. 

460. If a sample of NaOH contains 2.00 per cent by weight of Na 2 CO 3 
and 6.00 per cent by weight of H 2 O and if 40.0 grams are dissolved in water 
and diluted to a liter, what is the normality of the resulting solution as a 
base? Assume complete neutralization. 

Ans. 0.935 N. 

461. If 50.00 grams of a solid dibasic acid (mol. wt. 126.0) are mixed with 
25.00 grams of a solid monobasic acid (mol. wt. 122.0) and the mixture is 
dissolved and diluted to 2,500 ml., what is the normality of the solution as 
an acid? 

Ans. 0.3995 N. 

462. What is the normality as an acid of a solution made by mixing the 
following components? Assume no change in volume due to chemical effects. 



NEUTRALIZATION METHODS 163 

(a) 160 ml. of 0.3050 N HC1, (6) 300 ml. of half-molar H 2 SO 4 , (c) 140 ml. 
containing 1.621 grams of IIC1, (d) 200 ml. containing 1.010 grams of II 2 SO 4 . 
Ans. 0.517 N. 



463. In preparing an alkaline solution for use in volumetric work, a student 
mixed exactly 46.32 grams of pure KOH and 27.64 grams of pure NaOH and, 
after dissolving in water, diluted the solution to exactly one liter. How many 
milliliters of 1.022 N HC1 are necessary to neutralize 50.00 ml. of the basic 
solution? 

464. One gram of a mixture of 50.00 per cent anhydrous sodium carbonate 
and 50.00 per cent anhydrous potassium carbonate is dissolved in water and 
17.36 ml. of 1.075 N acrid are added. Is the resulting solution acid or alkaline? 
How many milliliters of 1.075 N acid or base will have to be added to make 
the solution exactly neutral? 

465. What would be the approximate normality of an acid solution made 
by mixing the following amounts of II 2 SO 4 solutions? (a) 160 ml. of 0.3050 
N solution, (6) 300 ml. of 0.4163 molar solution, (c) 175 ml. of solution con- 
taining 22.10 grams H 2 SO 4 , (d) 250 ml. of solution (sp. gr. 1.120, containing 
17.01 per cent H 2 SO 4 by weight). 

466. What is the normality of an alkali solution made by mixing 50.0 ml. 
of a solution containing 5.00 grains of NaOH with 100 ml. of a solution con- 
taining 2.90 grams of Ba(OH) 2 .8II 2 O and diluting with water to 250 ml.? 

70. Volume-normality-milliequivalent Relationship. A normal 
solution contains one gram-equivalent of solute per liter of solu- 
tion, or one gram-milliequivalent weight per milliliter of solution. 
It follows that the product of the number of milliliters of a given 
solution and the normality of the solution must give the number 
of milliequivalents of solute present, or 

ml. X N = number of milliequivalents 

where ml. = volume, milliliters 

N = normality 

This simple relationship is the basis of most calculations involving 
simple volume relationships between solutions and is illustrated 
in the following sections. 

71. Adjusting Solution to a Desired Normality. A solution 
with a given normality is often found to be too concentrated or 
too dilute for the purpose for which it is to be used. In order to 
decrease its concentration, water is usually added; and in order 
to increase its concentration, a solution is added which contains 
the solute in greater concentration than the one given. The 



164 CALCULATIONS OF ANALYTICAL CHEMISTRY 

amounts required in each case may be determined by simple 
calculation. 

EXAMPLE I. To what volume must 750.0 ml. of a 2.400 normal 
solution be diluted in order to make it 1.760 normal? 
SOLUTION : 

Before diluting, number of gram-milliequivalents 

= 750.0 X 2.400 = 1,800 
After diluting to x ml., these would be 

1,800 gm.-milliequivalents in x ml. 

= normality = 1.760 
x 

Solving, 

x = 1,023 ml. Ans. 

EXAMPLE II. How much 0.600 normal base must be added 
to 750 ml. of a 0.200 normal base in order that the resulting solu- 
tion shall be 0.300 normal? 
SOLUTION : 

Let x = milliliters of 0.600 N base added 
Total volume after dilution = 750 + x 
Total number of gram-milliequivalents present = 

(750 X 0.200) + (O.GOOr) 
Resulting normality (number of gram-milliequivalents per 

milliliter) = 0.300 
(750 X 0.200) + O.GOOx _ 

- frA - i - - U. 



fr 

750 + x 

= 250 ml. Ans. 

Problems 

467. Each milliliter of a solution of sodium carbonate contains exactly 
0.0109 gram of pure Na 2 CO 3 . To what volume must 100 ml. of the solution 
be diluted to make it exactly N/100? 

Ans. 2,056 ml. 

468. What volumes of 3.00 N and 6.00 N hydrochloric acid must be mixed 
to make a liter of 5.00 N acid? 

Ans. 667 ml. 6 N, 333 ml. 3 N. 

469. A solution of sulfuric acid is standardized gravimetrically, and it is 
found that 25.00 ml. will precipitate 0.3059 gram of BaSO 4 . To what volume 
must a liter of the acid be diluted in order to be exactly N/10? 

Ans. 1,047 ml. 



NEUTRALIZATION METHODS 165 

470. A solution of sodium hydroxide is found on analysis to be 0.5374 
normal, and a liter of it is available. How many milliliters of 1.000 N NaOH 
solution must be added in order to make the resulting solution 0.6000 normal? 

Ans. 156.5 ml. 

471. How much water must be added to 760 ml. of 0.2500 M barium 
hydroxide solution in order to prepare a tenth-normal solution? How many 
grams of Ba(OH)2.8H 2 O must be dissolved and diluted to 400 ml. to prepare a 
twelfth-normal solution? How many moles per liter, and how many gram- 
equivalent weights per liter does this last solution contain? 

Ans. 3,040 ml. 5.26 grams. 0.04167 mole, 0.08333 gram-equivalent 
weight. 

472. A 10-ml. pipetful of H 2 S0 4 (sp. gr. 1.80, containing the equivalent of 
80.0 per cent SO 3 by weight) is diluted to 500 ml. What is the normality of 
the solution as an acid? How many milliliters of 4.00 molar H 3 PO 4 should be 
added to this solution so that the resulting mixture will be 1.00 normal as an 
acid in reactions where neutralization to Na 2 SO 4 and Na 2 HPO 4 takes place? 

Ans. 0.719 N. 20.1ml. 

473. What volume of 0.2063 N KOH must be added to 150.0 ml. of 
0.1320 N KOH in order that the resulting solution shall have the same basic 
strength as a solution which contains 15.50 grams of Ba(OII) 2 per liter? 

Ans. 288.6 ml. 



474. What volumes of 0.500 N and 0.100 N HC1 must be mixed to give 2 
liters of 0.200 N acid? 

476. How many milliliters of water must be added to a liter of 0.1672 N 
sulfuric acid to make it exactly 0.1000 N? 

476. If 10.0 ml. of H 2 SO 4 (sp. gr. 1.50 containing the equivalent of 48.7 per 
cent of combined SOs by weight) are diluted to 400 ml., what is the normality 
of the solution as an acid? What volume of 6.00 molar IT 2 SO 4 should be added 
to this in order to make the resulting mixture 1.00 normal as an acid? 

477. If 10.0 ml. of Na 2 CO 3 solution (sp. gr. 1.080, containing 8.00 per cent 
Na 2 CO 3 by weight) are diluted to 50.0 ml, what is the normality of the result- 
ing solution as a base? What volume of 4.00 molar K 2 C0 3 solution should be 
added to this solution so that the resulting mixture will be 1.00 normal? 

478. A 500-ml. graduated flask contains 150.0 ml. of 0.2000 N sulfuric acid. 
By addition of more concentrated sulfuric acid, the solution is brought up to 
the mark and after mixing is found to be exactly 0.3000 N. What was the 
normality of the acid added? 

479. What volume of sulfuric acid (containing 0.150 gram of H 2 SO 4 per 
milliliter) should be added to 250 ml. of sulfuric acid (sp. gr. 1.035, con- 
taining 5.23 per cent H 2 S0 4 by weight) in order that the resulting solution 
shall be one-molar? 



166 CALCULATIONS OF ANALYTICAL CHEMISTRY 

480. A chemist desires to prepare approximately 14.00 liters of exactly 
0.5000 N NaOII. How many grams of the solid should be weighed out? 
After preparing 14.00 liters of the solution, the analyst standardizes it and 
finds it to be actually 0.4895 N. What volume of 6.00 N NaOH should be 
added to bring the solution up to half normal? After adding approximately 
this amount, the analyst standardizes the solution and finds it to be 0.5010 N. 
How much water should now be added? 

72. Volume and Normality Relationships between Reacting 
Solutions. Since a gram-milliequivalent weight of an acid will 
just neutralize a gram-mil liequivalent weight of a base and since 
the number of milliequivalents in each case is found by multiply- 
ing the number of millilitcrs of solution by its normality, we have 
the following simple relationship between two reacting solutions: 



ml.,i X N^ = ml.B X N# 

A solution can therefore be standardized by determining what 
volume of it will exactly react with a definite volume of another 
solution the normality of which is already known. The normali- 
ties of the two solutions will then be in inverse ratio to the respec- 
tive volumes used. Thus 50 ml. of any half-normal acid will 
neutralize 50 ml. of any half-normal base, or 100 ml. of quarter- 
normal base, since the solutions contain the same number of 
equivalent weights of reacting substance (i.e., 25 gram-milliequiva- 
lents). To neutralize 60 ml. of 0.5 N alkali solution (30 milli- 
equivalents), 15 ml. of 2 N acid (30 milliequivalents) will be 
required regardless of the chemical composition of the acid or 
alkali used. The chemical compositions of the reacting substances 
are taken into account in preparing their standard solutions. 

EXAMPLE. What is the normality of a solution of H 2 SO 4 if 
27.80 ml. are required to neutralize a 25-ml. pipetful of 0.4820 

N alkali? 
SOLUTION: 

27.80 X x = 25.00 X 0.4820 
x = 0.4334 N. Ans. 

Problems 

481. How many milliliters of normal sodium hydroxide solution are 
required to neutralize 5 ml. of (a) N HC1, (6) N/2 IIC1, (c) N/5 H 2 SO 4 , 
(d) N/5 HC1? 

Ans. (a) 5 ml, (6) 2.5 ml., (c) 1 ml., (d) 1 ml. 



NEUTRALIZATION METHODS 167 

482. A solution of HC1 contains 0.1243 gram-equivalent of HC1 per liter. 
How many milliliters of half-normal KOH solution are necessary to neutralize 
10.00 ml. of the acid? 

Ans. 2.486 ml. 

483. A solution of H 2 SO 4 is 0.1372 normal. How many milliliters of 0.1421 
normal KOH solution are required to neutralize 13.72 ml. of the acid? 

Ans. 13.24 ml. 

484. Convert 42.95 ml. of 0.1372 normal hydrochloric acid to the equivalent 
volume of normal solution. 

Ans. 5.892 ml. 

486. Subtract 34.37 ml. of 0.1972 HC1 from 42.00 ml. of 0.2000 N HC1 by 
converting both values to the equivalent volumes of normal acid. Express the 
answer in terms of (a) milliliters of 1.000 N HC1, (6) number of milliequivalents 
of IIC1, (c) number of milliliters of 0.5000 N NaOII. 

Ans. (a) 1.622 ml, (6) 1.622, (c) 3.244 ml. 

486. To neutralize 10.00 ml. of dilute acetic acid, 13.12 ml. of 0.1078 N 
KOH were required. What was the normality of the acid? 

Ans. 0.1415 N. 

487. A solution containing 31.21 ml. of 0.1000 N HC1 is added to a solution 
containing 98.53 ml. of 0.5000 N H 2 SO 4 , and 50.00 ml. of 1.002 N KOH are 
added. Is the resulting solution acid or alkaline? How many milliliters of 
0.3333 N acid or alkali will make it exactly neutral? 

Ans. Acid. 6.85 ml. of alkali. 

488. If 50.00 ml. of 1.087 normal HC1 are added to 28.00 ml. of a solution 
of a solid substance having an alkaline reaction, the alkali is more than neu- 
tralized. It then requires 10.00 ml. of 0.1021 N alkali to make the solution 
exactly neutral. How many milliequivalents of base per milliliter did the 
original solution of solid substance contain, and what was its normality as an 
alkali? 

Ans. 1.904, 1.904 N. 



489. Given: Standard sulfuric acid = 0.1072 N 

Standard sodium hydroxide = 0.1096 N 

How many milliliters of the sodium hydroxide solution are equivalent to 
26.42 ml. of the sulfuric acid solution? 

490. If 50.0 ml. of 6.00 N ammonium hydroxide and 50.0 ml. of 6.00 N 
hydrochloric acid are mixed, what is the approximate normality of the resulting 
ammonium chloride solution? 

491. How many milliliters of 0.300 N II 2 SO4 will be required to (a) neutralize 
30.0 ml. of 0.500 N KOH, (b) neutralize 30.0 ml. of 0.500 N Ba(OH) 2 , (c) neu- 
tralize 20.0 ml. of a solution containing 10.02 grams of KHCO 3 per 100 ml., 
(d) give a precipitate of BaSO 4 weighing 0.4320 gram? 



168 CALCULATIONS OF ANALYTICAL CHEMISTRY 

492. 1.000 ml. NaOH o 1.012 ml. HC1 

HC1 = 0.4767 N 

If 100.0 ml. of the alkali have been diluted to 500.0 ml. with the idea of 
preparing an exactly N/10 solution, how much too large is the volume? 

73. Determination of the Normality of a Solution. A solution 
may be standardized (i.e., its normality may be determined) in 
a variety of ways. In a few specific cases, it is possible to pre- 
pare a standard solution by accurately weighing out the solute, 
dissolving, and diluting to a definite volume. This method is 
applicable only to solutions of such substances as can be weighed 
out accurately and the composition and purity of which are 
definitely known. 

In some cases, it is possible to determine the normality of a 
given solution by gravimetric methods, i.e., by taking a definite 
volume of solution and precipitating the principal constituent in 
the form of a weighable compound of known composition. From 
the weight of this compound the weight of the solute in the vol- 
ume of solution taken is calculated. This gives a direct measure 
of the normality. For example, if a certain volume of hydrochloric 
acid is treated with an excess of silver nitrate, the weight of the 
precipitated silver chloride is a measure of the weight of hydrogen 
chloride in a liter of the acid. Since a liter of normal hydrochloric 
acid contains 36.47 grams of HC1, the normality of the solution 
is found by direct proportion. 

A solution is most often standardized, however, by determining 
the exact volume of it required to react with a known weight of 
substance of known purity (usually, but not necessarily, 100 per 
cent pure). One liter of a normal solution of an acid, for example, 
contains one gram-equivalent weight of that acid and therefore 
must just neutralize one gram-equivalent weight of any base, or 
one milliliter (a more convenient unit for ordinary experimental 
work) of the acid will neutralize one gram-milliequivalent weight 
of *any base. One milliliter of normal acid will just neutralize 
one milliequivalent weight in grams of any base. For example, 
it will neutralize Na 2 C0 3 /2,000 = 0.05300 gram of pure sodium 
carbonate, K2CO3/2,000 = 0.06910 gram of pure potassium car- 
bonate, or NaOH/ 1,000 = 0.04000 gram of pure sodium hydroxide. 
If 1 ml. of an acid solution were found to neutralize 0.1060 gram 



NEUTRALIZATION METHODS 169 

(i.e., 2 gram-milliequivalents) of pure sodium carbonate, the acid 
would be two-normal. If 1 ml. of an acid solution were found to 
neutralize 0.02000 gram (}^ gram-milliequivalent) of pure sodium 
hydroxide, the acid would be one-half normal. The same reason- 
ing holds true for the standardization of alkali solutions against 
acids and, as will be seen later, for the standardization of solu- 
tions of oxidizing, reducing, and precipitating agents. In calcu- 
lating the normality of a solution standardized in this way, the 
number of grams of pure standardizing agent divided by its milli- 
equivalent weight gives the number of gram-milliequivalents pres- 
ent. This must be the same as the number of gram-milliequivalents 
of substance in the solution used. Since this equals the number 
of milliliters times the normality, 

ml. 8 X N. = 

or 

__ grams* 
3 ml., X e x 

where e x is the milliequivalent weight of pure substance x which 
is titrated with solution s. 

EXAMPLE. A sample of pure oxalic acid (H2C 2 O4.2H 2 O) weighs 
0.2000 gram and requires exactly 30.12 ml. of potassium hydroxide 
solution for complete neutralization. What is the normality of 
the KOH solution? 

SOLUTION : The milliequivalent weight of oxalic acid is 



200 

The number of milliequivalents of oxalic acid present is 
0.2000/0.06303. The number of milliequivalents of KOH re- 
quired is 30.12 X N. 

OA1x, XT 0.2000 

30 ' 12XN== 006303 

N = 0.1053. Ans. 

74. Conversion of Data to Milliequivalents. In general, the 
student will usually find that the most satisfactory initial step in 
solving problems in analytical chemistry is to convert amounts of 



170 CALCULATIONS OF ANALYTICAL CHEMISTRY 

reacting substances to the corresponding number of milliequiva- 
lents of these substances. Since the number of milliequivalents 
of reacting substances are the same, such problems resolve them- 
selves into the simplest types of algebraic equations. The fol- 
lowing three formulas are of general applicability. 

1. Solution s of given normality: 

ml., X N, = no. of me.-wts. of solute 

2. Solution s of given specific gravity and percentage composi- 
tion: 

percentage x in solution 
ml. 8 X sp. gr. s X j-QQ 

= no. of me.-wts. of solute 

e x 

3. Solid x : 

= no. of me.-wts. of solid 

e x 

where e x = gram-milliequivalent weight of solid or solute 

Even in gravimetric analysis, the chemical factor (Sec. 52) ex- 
presses nothing more than a ratio between two equivalent or milli- 
equivalent weights. For example, the chemical factor 2Fe/Fe2O 3 
represents the weight of iron equivalent to a unit weight of ferric 

oxide. It is identical to the fraction ,.. ^ L r, n ^> which is the 

Jp 6203/2,000 

ratio of the milliequivalent weights of the two substances. 

Problems 

493. A hydrochloric acid solution is of such strength that 45.62 ml. are 
equivalent to 1.600 grams of pure Na 2 CO 3 . Calculate: (a) the number of gram- 
equivalents of Na 2 CO 3 neutralized by 1.000 liter of the acid, (6) the number 
of gram-milliequivalents of Na 2 CO 3 neutralized by 1.000 ml. of the acid, 
(c) the normality of the acid. 

Ans. (a) 0.6616, (6) 0.6616, (c) 0.6616 1ST. 

494. What is the normality of a solution of HC1 if 20.00 ml. are required to 
neutralize the NH 3 that can be liberated from 4 millimoles of (NH 4 ) 2 SO 4 ? 

Ans. 0.4000 N. 

496. How many milliliters of 3.100 N NaOH will be neutralized by (a) 105.0 
ml. of II 2 SO 4 (sp. gr. 1.050), (6) 10.50 grams of SO 3 ? 
Ans. 53.44 ml., 84.61 ml. 

496. Three millimoles of pure thiourea, CS(NH 2 ) 2 , are digested with con- 
centrated H 2 SO 4 and the nitrogen thereby converted to ammonium bisulfate. 



NEUTRALIZATION METHODS 171 

Excess NaOH is added and the liberated NII 3 is caught in 25.0 ml. of H 2 SO4 
(1.00 ml. - 2.00 ml. NaOH ^ 0.0315 gram H 2 C 2 O 4 .2H 2 O). The excess acid 
then requires 20.0 ml. of KOH. How many millimoles of P 2 O 6 would each 
milliliter of the KOH be equivalent to in the neutralization of H 3 PO 4 to the 
point of forming K 2 HPO4? 
Ans. 0.0813 millimoles. 

497. A 10-ml. pipetful of dilute sulfurie acid was standardized gravi- 
metrically by adding an excess of BaCl 2 , filtering, igniting, and weighing the 
resulting precipitate. The weight was found to be 0.2762 gram. Calculate 
the normality of the acid. 

Ans. 0.236G N. 

498. The normality of a sulfuric acid solution is 0.5278. If 38.61 ml. 
of the acid are equivalent to 31.27 ml. of a solution of NaOH, calculate the 
normality of the NaOH. If 38.61 ml. of the acid are equivalent to 62.54 ml. 
of a solution of Ba(OII) 2 , what is the normality of the Ba(OH) 2 ? 

Ans. 0.6516 N. 0.3258 N. 

499. Calculate the normality of a. solution of hydrochloric acid and of 
sodium hydroxide from the following data: 

1.000 ml. of H(a ~ 0.9492 ml. of NaOH 
39.81 ml. of HC1 o 0.6293 gram of AgCl 

Ans. IIC1 = 0.1105 N, NaOH = 0.1162 N. 

600. A sample of pure CaC0 3 weighs 1.0000 gram and requires 40.10 ml. 
of a solution of HC1 for neutralization. What is the normality of the acid? 
What volume of sulfuric acid of the same normality would be required for the 
same weight of CaCO 3 ? What volume of KOH solution of which 20.00 ml. 
will neutralize 1.420 grams of KHC 2 O 4 .H 2 O would be neutralized by 50.32 ml. 
of the acid? 

Ans. 0.4985 N. 40.10ml. 51.59ml. 

601. To a sample of sodium carbonate (99.20 per cent pure Na 2 C0 3 ) 
weighing 1.0500 grams are added 48.24 ml. of a solution of acid. This is in 
excess of the amount required for complete neutralization. The resulting 
solution is brought back to the neutral point with exactly 1.31 ml. of sodium 
hydroxide solution of which 1.000 ml. is equivalent to 1.010 ml. of the acid. 
Calculate the normality of the acid. 

Ans. 0.4189 N. 

602. In standardizing an alkali against 0.1200 gram of a solid acid (equiva- 
lent weight = 114.7), 38.92 ml of the alkali are added before it is realized 
that the end point has been overstepped. By introducing 0.0050 gram of 
pure H 2 C 2 04.2H 2 O into the solution, it is found that 0.58 ml. of the alkali is 
required to make the solution neutral. What is the normality of the alkali? 

Ans. 0.02850 N. 



172 CALCULATIONS OF ANALYTICAL CHEMISTRY 

603. A solution of sulfuric acid was standardized against calcium carbonate 
containing 91.90 per cent CaCOs and no other basic material. The sample 
weighing 0.7242 gram was titrated by adding an excess of acid (29.97 ml.), and 
the excess was titrated with 10.27 ml. of NaOH solution (1.000 ml. of the 
acid o 1.024 ml. of the NaOH). Calculate the normality of each solution. 

Ans. H 2 SO 4 = 0.6664 N. NaOH = 0.6507 N. 

604. A sample of pure potassium acid phthalate (a monobasic acid, 
KHC 8 H 4 O4) weighing 4.070 grams is titrated with NaOH solution and back- 
titrated with HCL NaOH required = 46.40 ml.; HC1 required = 5.35 ml. 
One milliliter HC1 ^ 0.01600 gram Na 2 O. How much water or how much 
6.00 N NaOH must be added to 500 ml. of the NaOH to bring it to 0.5000 N? 

Ans. 0.96 ml. of 6 N NaOH. 



606. 10.0 ml. NaOH o 0.0930 II 2 C 2 O 4 .2H 2 
1.00 ml. NaOH o 0.850 ml. HOI. 
What is the normality of the II Cl solution? 

606. What would be the normality of a solution of (a) HC1 and (6) H 2 SO 4 , 
if 40.0 ml. of the acid are required to neutralize 0.500 gram of pearl ash con- 
taining 95.0 per cent total alkali calculated as K 2 CO 3 ? 

607. Three millimoles of pure urea, CO(NH 2 ) 2 , are digested with concen- 
trated H 2 S0 4 and the nitrogen is thereby converted to ammonium bisulfate. 
Excess NaOH is added and the liberated NH 3 is caught in a 25-ml. pipetful of 
0.5200 N II 2 SO 4 . How many milliliters of NaOH (each milliliter will neutralize 
0.01640 gram H 2 C 2 O 4 .2H 2 O) will be required to neutralize the excess acid? 
How many millimoles of hydrated Al 2 Oa will each milliliter of the above II 2 SO 4 
be capable of reacting with to form A1 2 (SO 4 ) 8 ? 

608. An acid solution is prepared by dissolving 19.264 grams of pure 
KHC 2 4 .H 2 C 2 O 4 .2H 2 in water and diluting to exactly 900 ml. Fifty milliliters 
of this solution are neutralized by 35.00 ml. of KOH solution. What is the 
normality of each solution? 

609. 1.000 ml. NaOH o 0.0302 gram H 2 C 2 O 4 .2H 2 O 
1.000 ml. HC1 o 0.1123 gram BaC0 8 

What is the ratio by volume of HC1 to NaOH? How much solid NaOH 
must be added to 800 ml. of the alkali solution so that when the resulting 
solution is diluted to 1,000 ml. it will be 0.5000 N? How much water must 
be added to 1,000 ml. of the HC1 to make it 0.5000 N? What is the value of 
1.000 ml. of the original NaOH in terms of grams of benzoic acid (HC 7 H 6 O 2 )? 

610. What is the normality of a solution of KOH if 20.60 ml. are required to 
neutralize (a) 32.35 ml. of H 2 SO 4 (sp. gr. 1.160), (b) 1.000 gram of P 2 O 5 (forming 
K 2 HPO 4 )? 

611. How many milliliters of H 2 SO 4 (sp. gr. 1.105) will be neutralized by 
(a) 20.00 ml. of 2.680 N NaOH, (6) 5.100 grams of Fe 2 O 3 ? 

612. 1.000 ml. NaOH =0= 1.342 ml. HC1 
1.000 ml. HC1 o 0.0225 gram CaC0 3 



NEUTRALIZATION METHODS 173 

How much water must be added to 1,000 ml. of the sodium hydroxide 
solution to make it half normal? How much hydrochloric acid (sp. gr. 1.190, 
containing 37.23 per cent HC1 by weight) must be added to 1,000 ml. of the 
arid solution to make it half normal? 

613. Pure dry sodium carbonate weighing 0.1042 gram is dissolved in 
50.00 ml. of 0.1024 N sulfuric acid and the solution heated to boiling to expel 
the carbon dioxide liberated by the reaction. The solution is then titrated 
with 0.1120 N sodium hydroxide. What volume of the base is necessary to 
neutralize the solution? 

614. Calculate the normality of a solution of NaOH from the following 
data: 

Weight of potassium acid phthalate (KHC 8 H 4 04) = 4.119 grams. 
NaOH used = 42.18 ml. 

HC1 used - 3.10 ml. 
1.000 ml. HC1 =0= 0.02577 gram K 2 O 

What volume of 2.000 N NaOH or of water should be added to 750 ml. of 
the NaOH in order to bring it to 0.5000 N? 

76. Calculation of Percentage Purity from Titration Values. 

Just as the normality of a solution can be found from the volume 
required to react with a definite weight of substance of known 
purity, the percentage purity of a substance can be determined 
from the volume of a solution of known normality required to 
react with a definite weight of the substance. For example, one 
milliliter of normal alkali solution will neutralize one milliequiva- 
lent weight in grams of any acid. If an acid is titrated with 
normal alkali and exactly two milliliters of the latter are required, 
it follows that two gram-milliequivalent weights of the acid must 
be present. If two milliliters of two normal alkali are required, 
then four gram-milliequivalent weights of the acid must be present. 
Jri other words, the number of milliliters multiplied by the normal 
value of the solution will give the number of milliequivalents (in 
grams) of substance reacted upon. The number of gram-milli- 
equivalents thus found multiplied by the milliequivalent weight 
of the substance reacted upon will give the number of grams of 
that substance. If the percentage is desired, all that is necessary 
is to divide this weight by the weight of sample taken and multiply 
by 100. 

In general, therefore, if a substance x requires a certain number 
of milliliters of a solution s of normality N and if e is the milli- 
equivalent weight of the substance, 



174 CALCULATIONS OF ANALYTICAL CHEMISTRY 

ml., X N* X e x = grams* 
and 

ml, XN, X e x ^ inn , 

iTr n r - r X 100 = per cent* 
Weight oi sample 

EXAMPLE I. A sample of soda ash (impure Na 2 CO3) is titrated 
with half -normal sulfuric acid. If the sample weighs 1.100 grams 
and requires 35.00 ml. of the acid for complete neutralization, 
what is the percentage of Na^COs in the ash, assuming no other 
active component to be present? 

SOLUTION: One milliliter of normal acid will neutralize one 
gram-milliequivalent weight of any base. Thirty-five milliliters 
of 0.5000 N acid will neutralize 35.00X0.5000= 17.50 gram- 
milliequivalent weights of any base. Since the milliequivalent 
weight of Na 2 CO 3 is Na 2 CO 3 /2,000 - 0.05300, 35.00 ml. of the 
0.5000 N sulfuric acid will react with 

35.00 X 0.5000 X 0.05300 - 0.9275 gram of Na 2 C0 3 

As this weight is contained in a sample weighing 1.100 grams, the 
percentage of Na 2 CO 3 in the sample is 



^77^ X 100 = 84.32 per cent. Ans. 
l.luu 

It is important to remember that the normality of a solution 
merely expresses the ratio of its concentration to that of a solu- 
tion containing one gram-equivalent weight of solute per liter 
(i.e., a normal solution). Consequently, if the normality of a 
solution is known, the value of a definite volume of it in terms 
of other elements, compounds, or radicals can be found directly, 
even though the solution may not be capable of reacting directly 
with these elements, compounds, or radicals. Thus, the weight 
of hydrogen chloride in 10.00 ml. of 0.1000 N hydrochloric acid is 



10.00 X 0.1000 X TT^ = 0.03G47 gram 



The weight of silver chloride precipitated by adding an excess of 
silver nitrate to 10.00 ml. of 0.1000 N hydrochloric acid is 



10.00 X 0.1000 X ~ = 0.1433 gram 



NEUTRALIZATION METHODS 175 

The weight of silver sulfate equivalent to the silver in the silver 
chloride precipitated by adding an excess of silver nitrate to 
10.00 ml. of 0.1000 N hydrochloric acid is 



10.00 X 0.1000 X ~ = 0.1559 gram 



The weight of barium in the barium sulfate obtained by adding 
an excess of barium chloride to the silver sulfate above is 



10.00 X 0.1000 X o-^ = 0.06868 gram 



In other words, as in the case of gravimetric computations, it is 
not necessary to calculate the weights of the intermediate product 
of a reaction. From the milliequivalent weight of the substance 
required, the weight of that substance can be determined directly. 
EXAMPLE IT. Given the same conditions as in Example I, 
what would be the percentage of CO 2 in the soda ash? 
SOLUTION: 

35.00 X 0.5000 = 17.50 gram-milliequivalents of CO 2 



17.50 X ~ = 0.3850 gram of C0 2 
z,uuu 

X 10 = 35 - P 61 " cent C 2 - 



EXAMPLE III. A 0.3000-gram sample of impure magnesium ox- 
ide is titrated with hydrochloric acid solution of which 3.000 ml. =0= 
0.04503 gram CaCO 3 . The end point is overstepped on the ad- 
dition of 48.00 ml. of the acid, and the solution becomes neutral 
on the further addition of 2.40 ml. of 0.4000 N sodium hydroxide. 
What is the percentage of MgO in the sample? 
SOLUTION: 

1 ml. HC1 =0= ^55 ^ 0.01501 gram of CaCO 3 



Normality of HC1 = 

48.00 X 0.3000 = 14.40 milliequivalents of HC1 
2.40 X 0.4000 = 0.96 milliequivalents of NaOH 
14.40 - 0.96 = 13.44 = net milliequivalents 

13 " 4 72 ' 000 X MO - 90.33 per cent MgO. An,. 



176 CALCULATIONS OF ANALYTICAL CHEMISTRY 

76. Volumetric Indirect Methods. Instead of titrating a sub- 
stance directly with a standard solution, it is frequently more 
feasible to allow the substance to react with a measured amount 
of a given reagent and then to titrate that part of the reagent 
left over from the reaction. This is an indirect method and is 
characterized by the fact that, other factors being fixed, a greater 
degree of purity of the sample corresponds to a smaller buret 
reading. 

In acidimetry and alkalimetry an outstanding example of an 
indirect method is the Kjeldahl method for determining nitrogen 
in organic material. The sample is digested with concentrated 
H 2 S04 in the presence of a catalyst and the nitrogen in the ma- 
terial thus converted to ammonium bisulfate. The resulting solu- 
tion is made alkaline with NaOH and the liberated ammonia gas 
distilled (through a condenser) into a measured volume of standard 
acid (NH 3 + H+ NH4 + ). The acid remaining in the receiving 
flask, after all the NH 3 has been liberated, is then titrated with 
standard NaOH solution. 

Calculation of a volumetric indirect method is usually best made 
by determining the total number of milliequivalents of reagent 
added, and subtracting the number of milliequivalents used in 
the titration. This difference is the number of milliequivalents 
of desired substance. 

EXAMPLE. A sample of meat scrap weighing 2.000 grams is 
digested with concentrated sulfuric acid and a catalyst. The re- 
sulting solution is made alkaline with NaOH and the liberated 
ammonia distilled into 50.00 ml. of 0.6700 N H 2 SO 4 . The excess 
acid then requires 30.10 ml. of 0.6520 N NaOH for neutralization. 
What is the percentage of nitrogen in the meat? 
SOLUTION: 

Milliequivalents of H 2 SO 4 = 50.00 X 0.6700 = 33.50 
Milliequivalents of NaOH = 30.10 X 0.6520 = 19.62 
Net milliequivalents = 33.50 - 19.62 = 13.88 

Since, in the above process, NH 3 + H+ > NH4 + , the milliequiva- 
lent weight of NH 3 is NH 3 /1,000 and that of nitrogen is N/1,000. 

13.88 X N/1,000 vx inA n . . , 

' ' X 100 = 9.72 per cent. Ans. 



NEUTRALIZATION METHODS 177 

Problems 

616. Calculate the percentage of carbon dioxide in a sample of calcium 
carbonate from the following data: 

Total volume of 0.5000 N HC1 = 35.00 ml. 
Total volume of 0.1000 N NaOH = 17.50 ml. 

Weight of sample = 1.000 gram 
Ans. 34.65 per cent. 

616. Given the following data, calculate the percentage purity of a sample 
of cream of tartar (KHC 4 II 4 O 6 ): 

Weight of sample = 2.527 grams 
NaOH solution used = 25.87 ml. 
II 2 S0 4 solution used = 1.27 ml. 
1.000 ml. of H 2 SO 4 - 1.120 ml. of NaOH 
1.000 ml. of H 2 SO 4 ~ 0.02940 gram CaCO 3 
Ans. 95.50 per cent. 

517. A sample of pearl ash (technical grade of K 2 C0 3 ) weighing 2.000 grams 
is titrated with HC1, requiring 25.00 ml. What is the alkaline strength of the 
ash in terms of per cent of K 2 O if 20.00 ml. of the HC1 will just neutralize the 
NH 3 that can be liberated from four millimoles of (NH 4 ) 2 HPO 4 ? 

Ans. 23.55 per cent. 

518. Calculate the percentage of K 2 CO 3 in a sample of pearl ash from the 
following data: 

Weight of sample = 2.020 grams 

IIC1 used = 49.27 ml. 
NaOH used = 2.17 ml. 
1.000 ml. HC1 =0= 0.02926 gram CaCO 3 
NaOH = 0.3172 N 

Ans. 96.25 per cent. 

519. Given four 10.00-ml. portions of 0.1000 normal hydrochloric acid 
solution, (a) How many grams of pure sodium carbonate will be neutralized 
by one portion? (&) How many grams of K 2 O are contained in that weight 
of potassium hydroxide neutralized by another portion of the acid? (r) A 
sample of calcium carbonate is decomposed by a portion of the acid. Calcu- 
late the weight of CaCO 3 decomposed, the weight of CO 2 liberated, and the 
weight of CaCl 2 formed, (d) Calculate the weight of KHC 2 4 .H 2 C 2 O 4 .2H 2 O 
equivalent in acid strength to a portion of the HC1. 

Ans. (a) 0.05300 gram. (6) 0.04710 gram, (c) 0.05004 gram, 0.02200 
gram, 0.05550 gram, (d) 0.08473 gram. 

620. Strong KOH will liberate NH 3 from ammonium salts. The liberated 
ammonia can be distilled and determined by absorbing it in standard acid 
and titrating the excess acid with standard alkali. From the following data, 
calculate the percentage of NH 3 in a sample of impure ammonium salt: 



178 CALCULATIONS OF ANALYTICAL CHEMISTRY 

Sample = 1.009 grams 
Standard acid used = 50.00 ml. of 0.5127 N 
Standard alkali required = 1.37 ml. of 0.5272 N 
Ans. 42.06 per cent. 

521. Rochelle salt is KNaC 4 H 4 O 6 .4II 2 O and on ignition is converted to 
KNaCO 3 . The original sample of 0.9546 gram is ignited and the product 
titrated with sulfuric acid. From the data given, calculate the purity of the 
sample: 

H 2 SO 4 used = 41.72 ml. 
10.27 ml. H 2 SO 4 * 10.35 ml. NaOH 

NaOlI = 0.1297 N 

NaOH used in titrating excess acid = 1.91 ml. 
. Ans. 76.95 per cent. 

522. A sample of zinc oxide is digested with 50.00 ml. of normal sulfuric 
arid. The excess acid is titrated with 2.96 ml. of 0.1372 normal alkali. The 
weight of sample is 2.020 grams. Calculate the percentage of purity of the 
sample. 

Ans. 99.89 per cent. 

623. If all of the nitrogen in 10.00 millimoles of urea, CO(NH 2 )2, is converted 
by concentrated H 2 SO 4 into ammonium bisulfate and if, with excess NaOH, 
the NIT 3 is liberated and caught in 50.00 ml. of HC1 (1.000 ml. ^ 0.03000'gram 
CaCOs), how much NaOH solution (1.000 ml. =0= 0.03465 gram H 2 C 2 O4.2H 2 O) 
would be required to complete the titration? 

A?is. 18.18 ml. 

524. The percentage of protein in meat products is determined by multi- 
plying the percentage of nitrogen as determined by the Kjeldahl method by 
the arbitrary factor 6.25. A sample of dried meat scrap weighing 2.000 grams 
is digested with concentrated H 2 SO4 and mercury (catalyst) until the nitrogen 
present has been converted to ammonium bisulfate. This is treated with 
excess NaOH and the liberated NH 3 caught in a 50-ml. pipetful of H 2 SO 4 
(1.000 ml. =c=0.01860 gram Na 2 O). The excess acid requires 28.80 ml. of 
NaOH (1.000 ml. ^ 0.1266 gram potassium acid phthalate, KHC 8 H 4 O 4 ). 
Calculate the percentage of protein in the meat scrap. 

Ans. 53.11 per cent. 

525. A sample of milk weighing 5.00 grams is digested with concentrated 
H 2 SO 4 (plus a catalyst) which converts the protein nitrogen in the milk to 
ammonium bisulfate. Excess NaOH is added and the liberated NH 3 is evolved 
and caught in 25.0 ml. of dilute H 2 SO 4 . The excess acid then requires 28.2 ml. 
of NaOH of which 31.0 ml. are equivalent to 25.8 ml. of the dilute H 2 SO 4 . 

The acid and base are standardized by evolving the NH 3 from 1.00 gram of 
pure NH 4 C1, passing it into 25.0 ml. of the above dilute H 2 SO 4 and titrating 
the excess acid with the above NaOH. A volume of 11.3 ml. of the NaOH is 
required. 



NEUTRALIZATION METHODS 179 

The arbitrary factor for converting nitrogen in milk and milk products to 
protein is 6.38. Calculate the percentage of protein in the above sample of 
milk. 

Ans. 3.30 per cent. 



526. From the following data, calculate the percentage purity of a sample 
of KHSO 4 : 

1.000 ml. HC1 o 1.206 ml. NaOH 
1.000 ml. HC1 o 0.02198 gram Na 2 CO 3 

Sample = 1.21 18 grams 
HC1 used = 1.53 ml. 
NaOH used = 26.28 ml! 

627. A sample of Rochelle salt (KNaC 4 H 4 O6.4H 2 0), after ignition in 
platinum to convert it to the double carbonate, is titrated with sulfuric acid, 
methyl orange being employed as an indicator. From the following data, 
calculate the percentage purity of the sample: 

Weight of sample = 0.9500 gram 
H 2 SO 4 used = 43.65 ml. 
NaOH used = 1.72 ml. 
1.000 ml. II 2 SO 4 ^ 1.064 ml. NaOH 
NaOH =0.1 321 N 

528. A sample of milk of magnesia [suspension of Mg(OH) 2 j weighing 5.000 
grams is titrated with standard HNO 3 , requiring 40.10 ml. What is the per- 
centage of MgO in the sample if 20.11 ml. of the HNO 3 will just neutralize the 
NH 3 that can be liberated from five millimoles of (NH 4 ) 3 AsO 4 .5H 2 O? 

529. The saponification number of a fat or oil is defined as the number 
of milligrams of potassium hydroxide required to saponify one gram of the 
fat or oil. To a sample of butter weighing 2.010 grams are added 25.00 ml. 
of 0.4900 N KOH solution. After saponification is complete, 8.13 ml. of 
0.5000 N HC1 solution are found to be required to neutralize the excess alkali. 
What is the saponification number of the butter? 

530. Samples of oxalic acid mixed with inert matter are given out for 
student analysis to determine by acidimetric titration the acid strength in 
terms of percentage H 2 C 2 O 4 .2H 2 O. However, a sample of pure potassium 
acid tartrate KHC 4 H 4 O 6 , is included among the samples. What percentage 
of H 2 C 2 O 4 .2H 2 O would the student report in this case? 

631. A sample of vinegar weighing 10.52 grams is titrated with standard 
NaOH. The end point is overstepped, and the solution is titrated back with 
standard HC1. From the following data, calculate the acidity of the vinegar 
in terms of percentage of acetic acid (HC 2 H 3 O 2 ) : 

NaOH used = 19.03 ml. 
HC1 used - 1.50 ml. 
1.000 ml. HC1 o 0.02500 gram Na*CO 8 
1.000 ml. NaOH o 0.06050 gram benzoic acid (C 6 H 6 COOH) 



180 CALCULATIONS OF ANALYTICAL CHEMISTRY 

632. The arbitrary factor 6.25 is used by agricultural chemists to convert 
percentages of nitrogen in meat products to percentages of protein. A sample 
of dried pork scrap is sold under a guarantee of a minimum of 70.00 per cent 
protein. A one-gram sample is digested with sulfuric acid and a catalyst, 
which converts all the nitrogen to ammonium bisulfate. Treated with excess 
NaOII, the ammonia is liberated and caught in a 25-ml. pipetful of H 2 SC>4 
(1.000 ml. o 0.02650 gram Na 2 CO 3 ). What is the maximum volume of 
0.5110 N NaOH required to titrate the excess acid if the sample conforms to 
the guarantee? 

77. Problems in Which the Volume of Titrating Solution Bears 
a Given Relation to the Percentage. In commercial laboratories 
where many similar titrations are made each day, it is often con- 
venient to simplify computation by taking each time for analysis 
a weight of sample such that the volume of standard solution 
used will bear some simple relation to the percentage of desired 
constituent. The advantages derived from such a procedure are 
the same as those discussed in Sec. 55, and the computations in- 
volved are similar in principle. In the volumetric problem, it is 
also possible to fix the weight of sample and determine the nor- 
mality of the titrating solution which must be used to fulfill a 
similar condition, although this type of problem is less often met 
with in practice. It is easier in practical work to vary a sample 
weight than it is to vary a solution concentration. In cither case, 
however, the required weight of sample or normality of solution 
is best found by directly applying the formula previously de- 
rived, viz., 

ml.. X N 8 X e x 



Weight of sample 



X 100 = per cent* 



In this type of problem, it will always be found that, of the five 
factors involved, two will be known and a ratio will be given be- 
tween two others, thus making possible the determination of the 
fifth factor. 

EXAMPLE. What weight of soda ash should be taken for analy- 
sis such that the percentage of Na 2 O present may be found by 
multiplying by 2 the number of milliliters of 0.2000 N acid solu- 
tion used in the titration? 
SOLUTION: 

ml.. X N a X e x vv 1An , 

m . ., ,. r X 100 = per cent, 

Weight of sample 



NEUTRALIZATION METHODS 181 

In the problem given, N and e x are known. A relation also exists 
between the ml. and the per cent whereby 

ml. X 2 = per cent 
Upon substitution, 

ml. X 0.2000 X Na 2 0/2,000 

117 . i , T i x iuu = mi. a x & 

Weight ot sample 

0.2000 X 62.00/2,000 x 1QO ^ 2 
Weight of sample 

Weight of sample = 0.3100 gram. Ans. 

The same precautions should be taken in solving this type of 
problem as were emphasized in the examples in Sec. 55, namely, 
that a numerical difference exists between a statement such as 
"the number of milliliters is three times the per cent," and the 
statement "the per cent is found by multiplying the number of 
milliliters by 3." Thus, in the above example, the weight of soda 
ash to be taken so that each millilitcr of 0.2000 N acid shall equal 
% of 1 per cent of Na 2 O is found as follows: 

1 X 0.2000 X 0.03100 x IOQ ^ i 
Weight of sample 2 

Weight of sample = 1.240 grams. Ans. 

Problems 

633. A sample of oxalic acid is to he analyzed by titrating with a solution 
of NaOII that is 0.1000 N. What weight of sample should be taken so that 
each milliliter of NaOH will represent 3 2 of 1 P cr cent of IW^O^I^O? 

Ans. 1.261 grams. 

534. In the analysis of oxalic acid using a one-gram sample, what must be 
the normality of the alkali used for titration so that the buret reading will 
equal one-half the percentage of I^CaO^H^O? 

Ans. 0.3173 N. 

535. In the analysis of a sample of soda ash, what weight of sample should 
be taken so that the volume in milliliters of 0.4205 normal acid required for 
complete neutralization and the percentage of Na 2 CO 3 in the sample will be 
in the respective ratio of 8:17? 

Ans. 1.049 grams. 

636. A sample of a certain acid weighed 0.8250 gram and was titrated with 
0.2000 N alkali. After the purity of the sample was calculated in terms of the 
percentage of constituent A, it was found that the percentage obtained was 



182 CALCULATIONS OF ANALYTICAL CHEMISTRY 

just equal to the equivalent weight of A as an acid. What volume of titrating 
solution was used? 
Ans. 41.25 ml. 

637. A sample of limestone is titrated for its value as a neutralizing agent. 
A one-gram sample is always taken. What must be the normality of the 
titrating acid so that every 10 ml. will represent 4J^ per cent of the neutral- 
izing value expressed in terms of percentage CaO? 

Ans. 0.1605 N. 

638. Samples of pickling solution are to be analyzed volumetrically for 
acidity, and results are to be expressed in terms of milliliters H 2 SC>4 (sp. gr. 
1.84, containing 95.60 per cent H 2 SO 4 by weight). The specific gravity of 
the pickling solution is 1.270, and a 25-ml. pipetful is taken for analysis. 
(a) What must be the normality of the standard alkali so that each milliliter 
used will represent 0.100 ml. of the H 2 SO 4 ? (6) So that every 10.0 ml. will 
represent 1.00 per cent of pure H 2 SO4? 

Ans. (a) 3.59 N. (b) 0.647 N. 

639. What weight of soda ash must be taken for analysis so that by using 
0.5000 N IIC1 for titrating (a) the buret reading will equal the percentage of 
Na 2 O, (6) three times the buret reading will equal the percentage of Na 2 O, 
(r) every 3 ml. will represent 1 per cent Na 2 O, (d) each milliliter will represent 
3 per cent Na 2 O, (e) the buret reading and the percentage of Na 2 O will be in 
the respective ratio of 2:3? 

Ans. (a) 1.550 grams, (6) 0.5167 gram, (c) 4.650 grams, (d) 0.5167 gram, 
(e) 1.033 grams. 

640. What weight of calcite (impure CaCO 3 ) should be taken for analysis 
so that the buret reading will be 2^ times the percentage of Ca in the sample? 
The solution used for the titratiori is IINOa of which 2.00 nil. o 1.00 ml. 
Ba(OH) 2 solution =0= 0.0400 gram potassium acid phthalate (KHC 8 H 4 O4). 

641. What weight of soda ash (technical Na 2 CO 3 ) should be taken for 
analysis so that when titrated with HC1 [1.00 ml. o 2.00 ml. Ba(OH) 2 solu- 
tion o 0.0254 gram KHC 2 O 4 .H 2 C 2 O 4 .2II 2 O] the buret reading will be three- 
quarters of the percentage of Na 2 O in the ash? 

642. In the standardization of an acid, it was titrated against 1.000 gram 
of calcium carbonate (98.56 per cent pure). If 46.86 ml. of HC1 were added, 
the CaCOa dissolved, and the excess acid titrated with 5.21 ml. of NaOH 
solution of which 1.000 ml. =c= 0.7896 ml. HC1, calculate the weight of crude 
pearl ash to be taken for analysis so that each milliliter of this HC1 will 
represent 2.00 per cent K 2 O. 

643. If 1.500 grams of crude K 2 CO 3 are taken for analysis, what must be 
the strength of the HC1 used in order that the buret reading will indicate twice 
the percentage of K 2 O in the sample? 

644. A 2,000-gram sample of nitrogenous organic matter is digested with 
concentrated H 2 SO 4 and a catalyst until the nitrogen in the sample has been 
converted to NH 4 HSO 4 . By adding excess NaOH, NHa is liberated and is 



NEUTRALIZATION METHODS 183 

completely caught in a cold 5 per cent solution of boric acid. It is then 
titrated directly with standard HC1. What must be the value of each milliliter 
of the acid in terms of pure Na2COs if the buret reading is 2% times the per- 
centage of nitrogen in the material? 

646. A sample of quicklime is to be analyzed for CaO and CaCOs by 
titrating with 0.3572 N HC1. It is desired to start with a 10.0-gram sample,, 
mix with water, dilute, and take aliquot portions of such size (a) that when 
titrated with HC1 [phenolphthalein being used as an indicator, in which case 
only the Ca(OH) 2 is neutralized] the number of milliliters will represent directly 
the percentage of CaO and (6) that, when titrated by adding an excess of HC1, 
heating, and titrating back with NaOH of the same normality as the HC1, 
the net number of milliliters of HC1 used will represent directly the percentage 
of total calcium in terms of CaO. What portions should be taken? 

78. Determination of the Proportion in Which Components Are 
Present in a Pure Mixture. Problems involving the determina- 
tion from titration values of the proportion in which components 
are present in a pure mixture are identical in principle with the 
so-called double chloride problems of gravimetric analysis (see 
Sec. 57, Example II), and the same algebraic method of solution 
may conveniently be used. The same type of analysis may be 
applied equally well to methods of oxidation and reduction. 

As shown in Sec. 57, the precision of the result of an analysis 
of this type is usually less than that of the data given, and there 
is often a decrease in the number of significant figures that ma^ 
properly be retained in the numerical answer. 

EXAMPLE. If 0.5000 gram of a mixture of calcium carbonate 
and barium carbonate requires 30.00 ml. of 0.2500 N hydrochloric 
acid solution for neutralization, what is the percentage of each 
component? 
SOLUTION : 

Let x = number of grams of CaCO 3 
y = number of grams of BaCO 3 
Then 

(1) x + y = 0.5000 

Number of gram-milliequivalents of CaC0 3 present 

x x 

CaCOa/2,000 ~ 0.05004 

Number of gram-milliequivalents of BaC0 3 present = 

y y 

BaCO 3 /2,000 0.09869 



184 CALCULATIONS OF ANALYTICAL CHEMISTRY 

Number of gram-milliequivalents of HC1 used = 30.00 X 0.2500 
Therefore 

- 



Solving equations (1) and (2) simultaneously, 

x = 0.247 
y = 0.253 

247 
Percentage of CaCO 3 = ;rk/^s X 100 = 49.4 per cent 



0253 
Percentage of BaCO 3 = - X 100 = 50.6 per cent 



79. Analysis of Fuming Sulfuric Acid. Case A. An important 
titration is that involved in the analysis of fuming sulfuric acid 
(oleum). This substance may be considered to be a solution of 
sulfur trioxide, SO 3 , in hydrogen sulfate, H 2 SO 4 , and when no 
other component is present, the analysis is made by dissolving 
a weighed sample in water and titrating with standard alkali. 

EXAMPLE I. A sample of fuming sulfuric acid weighing 1.000 
gram when dissolved in water requires 21.41 ml. of 1.000 N NaOH 
solution for neutralization. What is the percentage of each com- 
ponent? 
SOLUTION : 

Method I. Since fuming sulfuric acid is a mixture of two pure 
components, the problem can be solved by the method of the 
preceding section. 

Let x = weight of free SO 3 

y = weight of H 2 S0 4 
x + y= 1.000 

SO^OOO + H 2 S0 4 /2,000 = 2L41 X l - 000 
When the simultaneous equations are solved, 

x = 0.222 gram S0 3 = 22.2 per cent 1 

y = 0.778 gram H 2 SO 4 = 77.8 per cent J S ' 

Method II. In dissolving the oleum, the SO 3 unites with part 
of the water to form H 2 SO 4 . If the total percentage of acid is 
computed in terms of H 2 S0 4 , the following result is obtained: 



NEUTRALIZATION METHODS 185 

21.41 X 1.000 X H 2 S0 4 /2,000 w t nn , n 
T- - X 100 = 105.0 per cent 

Since, in the original mixture, S0 3 + H 2 SO 4 = 100.00 per cent, 
the difference of 5.0 per cent is caused by the water which has 
combined with the SO 3 . The S0 3 and H 2 O combine mole for 

mole. 

SO 
5.0 X ^7-7! = percentage of 

X12VJ 



= 22.2 per cent SO 3 1 A 
= 77.8 per cent H 2 SO 4 J 



100.0 - 22.2 = 77.8 per cent H 2 SO 4 

Method III. In dissolving the oleum, the free SOs unites with 
water to form H 2 SO 4 (SO 3 + H 2 O -> II 2 S0 4 ). The percentage of 
total SOs (combined and free) in the original solution is found 
as follows: 

21.41 X 1.000 X SOs/2,000 

- - - - 



l.UUU 



v , nn 

X 100 = 85.70 per cent 



Since the original solution consisted of free SO 3 , combined SO 3 , 
and combined H 2 O, the percentage of combined II 2 in the solu- 
tion is 100.00 - 85.70 = 14.30 per cent. 

Percentage of original H 2 SO 4 = 14.30 X Jl ^ 

il 2 ij 

= 77.8 per cent ) , 
Percentage of free SO 3 = 22.2 per cent J 

Case B. Fuming sulfuric acid often contains small amounts 
of SO 2 which with water forms II 2 SO 3 and is included in the alkali 
titration: 

H 2 SO 3 + 2OH- - SO 8 - + 2H 2 O 



This is when phenolphthalein is used as the indicator. With 
methyl orange, the color change takes place at the bisulfite stage i 

H 2 S0 3 + OH~ -> HSOr + II 2 

In case S0 2 is present, its amount is usually determined in a 
separate sample by titration with a standard oxidizing agent, and 
the other components are then computed from the alkali titra- 
tion values in the usual way, with a correction for the volume of 
alkali used by the S0 2 . 



186 CALCULATIONS OF ANALYTICAL CHEMISTRY 

EXAMPLE II. A sample of fuming sulfuric acid containing 
H 2 S0 4 , 80s, and SO2 weighs 1.000 gram and is found to require 
23.47 ml. of 1.000 alkali for neutralization (phenolphthalein as 
indicator). A separate sample shows the presence of 1.50 per 
cent S0 2 . Find the percentages of S0 3 and H 2 S0 4 . 
SOLUTION: 



Volume of alkali used by SO 2 = LOQO = * 47 mL 



Volume of alkali used for H 2 S0 4 + S0 3 = 23.47 - 0.47 = 23.00 ml. 
Percentage of HaSCX + S0 3 = 100.00 - 1.50 = 98.50 per cent 

Let x = weight of SO 3 

y = weight of H2S04 
x + y = 0.9850 

<3 / mn + rr qr / /9 MM = 23.00 X 1,000 

fcUs/ &, \j\j\) xi2^^4/ ^IHJU 

Solving, 

x = 0.635 gram SO 3 = 63.5 per cent \ 
y = 0.350 gram H 2 SO 4 = 35.0 per cent J 

Problems 

546. A mixture consisting entirely of lithium carbonate and barium 
carbonate weighs 1.000 gram and requires 15.00 ml. of N HC1 for neutraliza- 
tion. Calculate the percentage of BaCO 3 in the sample. 

Ans. 71. 2 per cent. 

547. A mixture of pure lithium carbonate and pure strontium carbonate 
weighs 0.5280 gram and requires 19.82 ml. of 0.5060 N acid for neutralization. 
What is the percentage of Li 2 O and SrO in the sample? 

Ans. Li 2 O = 16.3 per cent, SrO = 41.8 per cent. 

548. What weight of barium carbonate must be added to 1.000 gram of 
lithium carbonate so that the mixture will require the same volume of standard 
acid for neutralization as would the same weight of pure calcium carbonate? 

Ans. 0.716 gram. 

549. A half-gram sample of a mixture of pure CaCOs and pure SrCO 3 
requires 30.00 ml. of 0.2726 N sulfuric acid for neutralization, (a) What 
would be the loss in weight of the original sample on strong ignition? (6) Cal- 
culate the combined weight of CaS0 4 and SrS0 4 obtained above, (c) What is 
the weight of CaCO 3 in the original sample? 

Ans. (a) 0.180 gram, (b) 0.647 gram, (c) 0.218 gram. 

650. The combined weight of LiOH, KOH, and Ba(OH) 2 in a mixture is 
.5000 gram, and 25.44 ml. of 0.5000 N acid are required for neutralization. 



NEUTRALIZATION METHODS 187 

The same amount of material with CO 2 gives a precipitate of BaCO 3 that when 
filtered is found to require 5.27 ml. of the above acid for neutralization. 
Calculate the weights of LiOH, KOH, and Ba(OH) 2 in the original sample. 
Ans. LiOH = 0.217 gram, KOH = 0.0567 gram, Ba(OH) 2 = 0.226 grain. 

661. A sample of fuming sulfuric acid, containing no SO 2 or other impurity, 
on titration is found to contain 108.5 per cent acid expressed in terms of H 2 SO 4 . 
Calculate the percentage of free SO 3 in the sample. 

Ans. 37.8 per cent. 

662. A sample of fuming sulfuric acid containing only SO 3 and H 2 SO4 
is titrated, and the percentage of total SO 3 (free and combined) is found to be 
84.00 per cent. What is the percentage of 1I 2 SO4 in the original sample? 

Ans. 87.1 per cent. 

663. A sample of fuming sulfuric acid containing only SO 3 and H 2 SO4 
weighs 1.4000 grams and requires 36.10 ml. of 0.8050 normal NaOH for 
neutralization. What is the percentage of each constituent in the sample? 

Ans. 91.98 per cent II 2 SO 4 , 8.02 per cent SO 3 . 

654. A solution of SO 3 in H 2 SO 4 requires 65.10 ml. of 0.9000 normal alkali 
for the titration of a sample weighing 2.604 grams. What is the proportion 
by weight of free SO 3 to H 2 SO 4 in the sample? 

Ans. 0.850. 

666. A sample of fuming sulfuric acid consisting of a solution of SO 3 and 
SO 2 in H 2 SO 4 is found to contain 2.06 per cent SO 2 . A sample weighing 1.500 
grams requires 21.64 ml. of 1.500 N KOH when phenolphthalcin is used as the 
indicator. What are the percentages of free SO 3 and H 2 SO 4 in the sample? 

Ans. 22.4 per cent free SO 3 , 75.6 per cent H 2 SO 4 . 



666. A mixture of pure sodium carbonate and pure barium carbonate 
weighing 0.2000 gram requires 30.00 ml. of 0.1000 N acid for complete 
neutralization. What is the percentage of each constituent in the mixture? 

567. A sample supposed to be pure calcium carbonate is used to standardize 
a solution of IIC1. The substance really was a mixture of MgCO 8 and BaCOs, 
but the standardization was correct in spite of the erroneous assumption. 
Find the percentage of MgO in the original powder. 

658. A mixture of BaCO 3 and CaCO 3 weighs 0.5000 gram. The mixture 
is titrated with HC1, requiring 12.90 ml. From the following data, calculate 
the percentage of barium in the mixture: 

30.40 ml. HC1 ~ 45.60 ml. NaOH 
2.000 ml. of NaOH will neutralize 0.07460 gram of NaHC 2 O 4 

669. Glacial acetic acid often consists of a mixture of two acids, namely, 
pure acetic acid, HC 2 H 3 O 2 or CH 3 COOH, and a small amount of acetic an- 
hydride, (CHjCO) 2 O. When dissolved in water, the anhydride forms acetic 



188 CALCULATIONS OF ANALYTICAL CHEMISTRY 

acid: (CH 3 CO) 2 O + H 2 O -> 2CH 3 COOH. A sample of the original substance 
weighing A grams is dissolved in water and requires B ml. of C normal NaOH 
for neutralization. Set up an expression showing how the percentage of acetic 
anhydride in the sample can be determined from this single titration. Express 
clearly the correct milliequivalents involved. 

660. A sample of P2Os is known to contain H 8 PO 4 as its only impurity. 
A sample is weighed in a closed container, the container is opened under water 
(P20 5 + 3H 2 O -* 2H 3 PO 4 ) and the solution is titrated with standard NaOH 
to form Na 2 HPO 4 at the end point. If A ml. of B normal NaOH were used, 
set up an expression to show how the number of grams of P 2 O& in the original 
mixture could be determined. Express all milliequivalent weights. 

661. The titration of a sample of fuming sulfuric acid containing no SO 2 
shows the presence of an equivalent of 109.22 per cent H 2 SO 4 . Calculate 
the percentage composition of the sample and the percentage of combined SO 3 . 

662. A sample of oleum weighing 1.762 grams requires 42.80 ml. of 0.8905 
N NaOH for neutralization. Calculate the proportion by weight of free SO 3 
to combined S0 3 in the sample. 

663. A mixture of SO 3 and H 2 SO 4 contains 91.18 per cent of total SO 3 . 
Calculate the volume of half-normal alkali required to titrate a solution of 
1.030 grains of the mixture. What is the percentage of free SO 3 in the mix- 
ture? What is the equivalent of the mixture in terms of aqueous sulfuric acid 
containing 93.19 per cent II 2 SO 4 ? 

664. A mixture of pure acetic acid and acetic anhydride is dissolved in 
water and titrated with NaOH. The acidity of the sample expressed in terms 
of HC 2 H 3 O 2 is found to be 114.0 per cent. What is the composition of the 
original mixture? Acetic anhydride reacts with water to form acetic acid: 
<CH 3 CO) 2 O + H 2 ~> 2HC 2 H 3 2 . 

80. Indicators. An indicator is used in volumetric analysis for 
the purpose of detecting the point at which a reaction is just com- 
pleted. The indicators used in acidimetry and alkalimetry are 
usually organic dyestuffs which are of one color in acid solution 
and of a distinctly different color in alkaline solution. They are 
usually in themselves either weak acids (e.g., phenolphthalein) or 
weak bases (e.g., methyl orange), and the change in color that they 
undergo can be attributed to the fact that the arrangement of 
the atoms in their molecules is somewhat different from the ar- 
rangement of the atoms in the molecules of their corresponding salts. 

Consider a weak organic acid indicator of the general formula 
HX. This acid ionizes as follows: HX +> H+ + X~. The undis- 
sociated molecule HX is, for example, colorless; the ion X~ is 
colored, usually because of a rearrangement of atoms to form a 
quinoid structure. In water solution the ionization of the acid 



NEUTRALIZATION METHODS 189 

is so slight that the color of the ion is too faint to be seen. The 
addition of an alkaline substance to the solution, however, by 
reacting with the hydrogen ion, displaces the above equilibrium 
to the right and increases the concentration of the X~ ion to the 
point where its color becomes visible. The ionization constant of 



the above indicator is as follows: mr^i = ^' anc ^ * s ca ^ e( ^ an 

[HAJ 

indicator constant. If it is assumed that with this type of indicator 
a titration is stopped when one half of the un-ionized molecules 
have been converted by a base to the colored ionic form, then 
[HX] = [X~] and the indicator constant is equal to the hydrogen- 
ion concentration of the solution at the end point. 

Similarly, a weak basic indicator of the general formula XOII 
ionizes as follows: XOH <= X + + Oil"", and the ionization constant 

(= indicator constant) is ^ rv L" LJ1 = K. In water solution the 

[AOHJ 

color of the XOH molecule predominates, but the addition of acid 
increases the concentration of the X~ form and the color changes. 
If it is assumed that the color change is seen when three-fourths 
of XOH has been converted to X~, then the hydroxyl-ion con- 
centration at the end point is equal to ^K. 

With a given concentration of indicator, the color change takes 
place at a point where the hydrogen-ion or the hydroxyl-ion con- 
centration in the solution has attained a definite value that is 
characteristic of the indicator in question. Thus, a solution con- 
taining about 0.001 per cent of phcnolphthalein turns from color- 
less to pink when the hydroxyl-ion concentration has attained the 
value of about 1 X 10~ 5 mole per liter, and the corresponding hy- 
drogen-ion concentration has therefore been reduced to about 
1 X 10~ 9 mole per liter (pH = 9). Figure 2 shows the approximate 
hydrogen-ion and hydroxyl-ion concentrations at which dilute 
solutions of the common indicators change color. It will be noted 
that on this chart each color change is spread over a certain range 
of pH values. Each indicator may be said to have a "range of 
doubt" over which the transition in shade of color is gradual; 
the average analyst might stop a titration anywhere within the 
range in question. 

81. Equivalence Point. The equivalence point in any titration 
is the point where the amount of titrating solution added is 



190 



CALCULATIONS OF ANALYTICAL CHEMISTRY 



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NEUTRALIZATION METHODS 191 

chemically equivalent to the amount of substance being titrated; 
the analyst attempts to make the end point (i.e., the point where 
the indicator changes color) coincide with this. In an acidimetric 
or alkalimetric titration the equivalence point is not necessarily 
the same as the neutral point (pH = 7). For example, in the 
titration of acetic acid with sodium hydroxide, when the latter 
has been added in an amount equivalent to the former, the 
acidity of the solution is the same as that resulting from dissolving 
the corresponding amount of sodium acetate in water. Such a 
solution is basic owing to hydrolysis of the salt. Similarly, in 
the titration of a weak base with a strong acid, the equivalence 
point is at a point where the solution is slightly acidic (pH < 7). 

Other conditions being equal, the correct indicator for a given 
titration is one of which the color change takes place when the 
solution has that pH value which exists in a solution obtained by 
dissolving in the same volume of water the salt formed by the 
neutralization. In other words, an indicator should be chosen 
that will change color at a pH value approximately equal to the 
pH value at the equivalence point. Just how that pH value can 
be determined is shown in the following section. 

82. Determination of pH Value at the Equivalent Point. In 
the following discussion, let us consider four general types of acidi- 
metric titrations: 

A . Titration of a strong acid with a strong base, or vice versa 

B. Titration of a weak acid with a strong base 

C. Titration of a weak base with a strong acid 

D. Titration of a weak acid with a weak base, or vice versa 
Let us also make use of the following symbols: 

pH = log JL = -log [H+] 
pOH = log = -log [OH-] 



Kw = ion-product constant of water = [H+][OH"~] = 1.0 X 
10- 14 (at 25C.) 

pW = log ^- = -log Kw - 14.0 (at 25C.) 
ivw 

Ka = ionization constant of the weak acid being titrated = 



192 



CALCULATIONS OF ANALYTICAL CHEMISTRY 



pA = log ==- = -log Ka 
Jva 

Kb = ionization constant of the weak base being titrated = 
[X+][OH-]/[XOH] 

pB = log = -log Kb 

C = molar concentration at the equivalence point, of the 
salt formed by the neutralization process 



20 Ail. N/2 solution diluted to 100 ml. 
| and titrated with N/2 solution 



HCl+NaOH 
HC 2 H 8 2 +NaOH- 



HC.H.0 9 fNH,QH 







n Phenol- 
LJphthalein 

f~1 Bromthymol 
LJblue 

Methyl 
red 

H Methyl 
1-1 orange 



12 16 20 24 28 
Titrating solution, ml. 

FIG. 3. Acidimetric titration curves. 

Case A Twenty milliliters N/2 HC1 diluted to 100 ml. with 
water and titrated with N/2 NaOIL 

At the beginning of the titration, [H+] = 20 X % mole per 
100 ml. = 0.10 molar. Therefore pH = 1.0. As the titration pro- 
gresses, the hydrogen-ion concentration decreases (pH increases), 
and as the equivalence point is approached the change becomes 
very rapid. At the equivalence point, the solution contains only 
sodium chloride dissolved in water, and, since there is no appre- 
ciable hydrolysis, pH = 7.0. Beyond the equivalence point, the 
solution is basic and the pH value rapidly drops to about 12. 
The graph for this titration, in which buret readings are plotted 
against corresponding pH values, is shown in Fig. 3, curve (A) (A). 
At the right of the figure are shown the approximate pH values 



NEUTRALIZATION METHODS 193 

at which four of the common indicators (phenolphthalein, brom- 
thymol blue, methyl red, and methyl orange) change color. So far 
as the titration of a strong acid [curve (A) (A)] or the titration 
of a strong base [curve (B) (B)] is concerned, it is seen that not 
only is the equivalence point at pH = 7.0, but near the equiva- 
lence point the change in pH is so rapid that any indicator chang- 
ing color between about pH = 3 and pH =11 should be suitable. 
In other words, in titrations of strong acids with strong bases, 
and vice versa, although an indicator changing at pH = 7 is in- 
dicated (e.g., brom thymol blue), yet the error involved in the 
use of such indicators as phenolphthalein or methyl orange is 
negligible being usually within the error of reading a buret. 

Case B. Twenty milliliters N/2 acetic acid (HC 2 H3O 2 ) diluted 
to 100 ml. and titrated with N/2 NaOH. 

At the beginning of the titration, the pH value is approximately 
3. This can be calculated from the ionization constant of acetic 
acid, Ka = 1.86 X 10~ 5 , thus: 

H0 2 II 3 2 <= H+ + C 2 H 3 2 - 
[H+][0 2 H 3 2 -] 



[HC 2 H 3 2 ] 
xXx 



= Ka= 1.86X 10- 6 
= 1.86 X 10- 6 



(0.10-z) 
x = 1.36 X 10~ 3 = [H+] 



- 2 - 87 



At the equivalence point, the pH value of the solution can be 
calculated from the following formula (which is general for titra- 
tions of this type) : 



This may be derived by considering the numerical equilibrium 
relationships at the equivalence point in the titration just cited. 
The salt, sodium acetate, formed at the equivalence point at con- 
centration C, hydrolyzes as follows: 

NaC 2 H 3 O 2 + H 2 + HC 2 H 3 O 2 + NaOH 
or 

C 2 H 3 O 2 - + H 2 O 4=i HC 2 H 3 O 2 + OH- 



194 CALCULATIONS OF ANALYTICAL CHEMISTRY 

The mass-action expression for this hydrolysis is 

m [HC 2 H 3 Q 2 ][OH-] 

w [C 2 H 3 2 -] 
but 

(2) [H+][OH-] - Kw 
and 

[H + ][C 2 H 3 2 -] 



[HC 2 H 3 2 ] 

Dividing (2) by (3) gives (1) 
Hence 

[HC 2 H 3 O 2 ][QH-] Kw 
[C 2 H 3 2 -] Ka 

but, as seen from the above hydrolysis equilibrium, 

[HC 2 H 3 2 ] = [OH-] 
and, if the extent of hydrolysis is not too great, 

[C 2 H 3 O-] = C (approx.) 
therefore, 

Kw 




Ka 

pH = -[log Kw (^2 log C + }/2 log Kw y% log Ka)] 
= y<i log Kw Yi log Ka + % log C 

= 1 A pw + y 2 P A + y 2 log c 

In the case at hand, we have Kw = 1.0 X 10~ 14 , Ka = 1.86 X 10"" 5 , 
and 

^ 0.010 mole NaC 2 H 3 O 2 



j - rr 1 
120 ml. solution 



~ U.Oooo molar 



. 
This is point C,on the titration curve (C) (C). 



NEUTRALIZATION METHODS 195 

The common indicator that changes color at approximately this 
point is phenolphthalein and is the indicator suitable for the titra- 
tion. As seen from the titration graph, the use of an indicator 
like methyl orange would give erroneous results. 

Case C Twenty milliliters of N/2 NH 4 OH diluted to 100 ml. 
and titrated with N/2 HC1. 

At the beginning of the titration, pH = 11.1, as calculated from 
the ionization constant of NH 4 OH (Kb = 1.75 X 10~ 5 ). At the 
equivalence point, the pH value can be calculated from the for- 
mula (which is general for titrations of this type) : 



This equation is derived from the hydrolysis constant of ammo- 
nium chloride by a method analogous to that in Case B. 

In the case at hand, we have Kw = 1.0 X 10~ 14 , Kb = 1.75 X 
10~ 5 , and C = 0.0833. 

14.0 4.76 (-1.08) 
- 2 2~~ 

= 5.16 

This is point D on the titration curve (D) (D). 

The common indicator that changes color at approximately this 
point is methyl red. As seen from the chart an indicator like phe- 
nolphthalein would give erroneous results. The change of color 
would be gradual, and the end point would occur considerably 
before the true equivalence point. 

Case D. The titration curves for the neutralization of a weak 
acid like acetic acid with a weak base like ammonium hydroxide 
and for the neutralization of a weak base with a weak acid are 
represented by curves (E) (E) and (F) (F) in the accompanying 
figure. The pH value at the equivalence point can be found by 
substituting in the following formula (which is general for titra- 
tions of this type) : 



Such titrations are of no value in general analytical work, for, as 
seen from the chart, there is no sudden inflection of the curve 



196 CALCULATIONS OF ANALYTICAL CHEMISTRY 

at the equivalence point, and no indicator has a sharp enough 
change in color to indicate the equivalence point with satisfactory 
precision. 
83. Calculation of the Degree of Hydrolysis of a Salt. Not 

only are the above formulas useful for calculating the pH value 
at the equivalence point in a given titration, but they can be used 
to calculate the approximate extent of hydrolysis of a salt of a 
weak acid or of a weak base. For example, in the hydrolysis of 
sodium acetate (C 2 H 8 O 2 " + H 2 - IIC 2 H 3 O2 + OH~), the value 
of the concentration of OH~ can be found from the pH value cal- 
culated from the appropriate formula above; the value of the con- 
centration of acetate is that of the concentration of the salt (0) 
in the formula. The ratio of [OII~] to [C2H 3 2 ~] indicates the 
degree of hydrolysis of the salt. 

EXAMPLE. What is the percentage hydrolysis of a 0.0010 molar 
solution of NH 4 C1 (NH 4 + + H 2 O -> NH 4 OH + 11+)? 
SOLUTION : 

pH = % pW - Y 2 pB - y z log C 
where pW =14 

pB = - log K NH4 oH = - log 1.75 X 10~ 5 

C = 0.0010 
Solving, 

pH = G.12 

[H+] = log 6l2 = 7 ' 6 X 1(H 

[NH 4 +] = 0.0010 

rjj+i 
Percentage hydrolysis = prf^-iTi x 10 = - 076 P er cent - 



Problems 

666. What is the pH value of a solution that at 25C. has a hydroxyl-ion 
concentration of 4.2 X 10~ 8 ? What color would be given to the solution by 
a drop of methyl orange? Of thymol blue? 

Ans. 6.62. Yellow. Yellow. 

666. What is the hydrogen-ion concentration of a solution that at 25C. 
has a pOH value of 8.85? What common indicator would change color at 
approximately this concentration? 

Ans. 7.08 X 1Q-*. Methyl red. 

667. A certain weak monobasic acid is colorless in acid solution and blue in 
alkaline solution. Assuming that the blue is seen when two-fifths of the 



NEUTRALIZATION METHODS 197 

indicator has been converted to ions and that at this point the pOH value of 
the solution is 3.6, what is the indicator constant of the indicator? 
Ans. 2.7 X 10- u . 

668. A certain weak monobasic acid has an ionization constant of 
2.0 X 10~ 4 . If 1/100 mole is dissolved in water and the solution diluted to 
200 ml. and titrated with 0.250 N NaOH, calculate the pH value of the solu- 
tion at the following points: (a) the original solution, (6) one-fifth of the way 
to the equivalence point, (r) at the equivalence point. 

Ans. (a) 2.50, (6) 3.10, (r) 8.16. 

569. What is the pH value of a 0.0100 molar solution of KCN? Of NH 4 C1? 
What common indicator is therefore suitable for the titratioii with HC1 of a 
solution approximately N/100 in NH 4 OH? 

Ans. 10.57. 5.62. Methyl red. 

570. What is the percentage hydrolysis at 25C. in a 0.0050 molar solution 
of potassium acetate? 

Ans. 0.032 per cent. 

671. What are the pH value and the percentage hydrolysis at 25C. in a 
0.010 molar solution of sodium formate (KncHOz = 2.1 X 10~ 4 )? 

Ans. 7.84, 0.0069 per cent, 

672. What is the percentage hydrolysis at 25C. in a 0. 10 molar solution 

of Na2C(MCQr + H 2 o <= iicor + on-)? 

A ns. 4.5 per cent. 

573. What are the pH value, the hydroxyl-ion concentration, and the per- 
centage hydrolysis at 25C. in a 0.10 molar solution of NaCN? 
Ans. 11.07, 1.1 X 10~ 3 , 1.1 per cent. 

674. How many moles per liter of KC1O are required to give a solution 
with a hydroxyl-ion concentration of 2.0 X 10" 6 at 25C.? 
Ans. 1.6 X 10~ 5 . 

576. A sample of vinegar weighing 6.00 grams is dissolved in water, diluted 
to 50.0 ml., and titrated with 0.505 N NaOH, phenolphthalein being used. 
After 12.40 ml. of the base have been added, it is found necessary to back- 
titrate with 2.00 ml. of 0.606 N HC1. What is the acidity of the vinegar in 
terms of percentage of acetic acid, HC 2 H 3 O 2 ? Assuming that this is the only 
acid present in appreciable amounts in the vinegar, calculate the pH value of 
the solution at the equivalence point at the end of the above titration. Is 
phenolphthalein shown to be suitable for this titration? 

Ans. 5.05 per cent. 8.82. Yes. 

676. If 400 ml. of a solution containing NH 4 OH are titrated with 0.250 N 
HC1, 40.0 ml. of the acid are required to reach the equivalence point. What 
is the pH value of the solution at the start of the titration, halfway to the 



198 CALCULATIONS OF ANALYTICAL CHEMISTRY 

equivalence point, and at the equivalence point? What indicator is thus 
shown to be suitable? 

Ans. 10.82, 9.24, 5.44. Methyl red. 

677. Formic acid (HCOOH) is a monobasic acid that is 4.6 per cent 
ionized in tenth-molar aqueous solution at 25C. Calculate the ionization 
constant of formic acid. If 50.0 ml. of 0.100 N HCOOH are diluted to 250 ml 
and titrated with 0.200 N NaOH, what would be the pH value at the equiva- 
lence point? What indicator is suitable for the titration? 

Ans. 2.13 X 10- 4 . 7.97. Cresol red. 



678. What is the pOH value of a solution the hydrogen-ion concentration 
of which at 25C. is 9.0 X 10~ 10 ? What common indicator would change 
color at approximately this concentration? 

579. What is the hydroxyl-ion concentration of a solution that at 25C. 
has a pH value of 6.30? What color would be given to the solution by a 
drop of congo red? Of cresol red? 

680. Derive the formula pH = J^ pW J/2 pB Yt, log C, which represents 
the pH value at the equivalence point in the titration of a weak base with a 
strong acid. 

581. A certain weak monoacidic organic base serves as an indicator. 
Assuming that the color change is seen when one-third of the indicator has 
been converted to ions and that at that point the pH value of the solution 
is 4.8, what is the indicator constant of the indicator? 

682. In the titration of a solution of a certain monoacidic base with HC1, 
with methyl red as the indicator, the appearance of a shade of pink in the 
solution is taken as the end point. On the assumption that the concentra- 
tion of the resulting salt is 0.100 N and that the indicator used is best suited 
for this titration, what is the approximate ionization constant of the base? 

683. Benzoic acid (CeHsCOOH) is a monobasic acid with an ionization 
constant of 6.6 X 10~" 6 . A sample of the pure acid weighing 0.610 gram is 
dissolved in 500 ml. of water and titrated with 0.500 N NaOH. Calculate the 
pH value of the solution at the start of the titration, halfway to the equivalence 
point and at the equivalence point. What indicator is shown to be suitable 
for this titration? Roughly sketch the titration graph. 

684. What is the percentage hydrolysis in a solution of 0.050 molar 
NaNO 2 ? 

686. By how many times is the percentage hydrolysis of NH 4 NO 8 increased 
when its 0.10 molar solution is diluted tenfold? 

686. What are the pH value and the percentage hydrolysis in a solution 
0.10 molar in KHCO 3 at 25C.? (HCO 3 - + H 2 O <=> H 2 CO, + OH~). 

687. How many grams of each of the following substances must be dissolved 
in 100 ml. in order that the resulting solution shall have a pH value of 9.0; 
(a) NH 8 , (6) NaOH, (c) KN0 2 , (d) NaHCO 3 ? 



NEUTRALIZATION METHODS 199 

588. What is the percentage hydrolysis in 0.10 M solution at 25C. of 
(a) Na 3 P0 4 (POr + H 2 O * HPO 4 ~ + OH-), (6) Na 2 HPO 4 (HPOr + H 2 * 
H 2 PO 4 ~ + OH-)? 

689. A certain organic amine is a monoacidic base like NtUOH and is 
soluble in water. Calculate its ionization constant from the fact that a tenth- 
molar solution of the base is 6.6 per cent ionized. What is the hydroxyl-ion 
concentration at the equivalence point in the titration of 200 ml. of a 0.20 
molar solution of the base with 0.500 N HC1? What indicator is suitable? 

690. A certain organic amine is a weak monoacidic base like NH 4 OH. Its 
ionization constant is 1.0 X 10~ 4 . If 100 ml. of a 0.020 molar solution is 
titrated with 0.020 N HC1, what is the hydroxyl-ion concentration at the 
equivalence point? Which of the following four indicators would be best 
suited for the titration: methyl orange, phenolphthalein, methyl red, brom- 
thymol blue? Carefully sketch the titration curve (pH against ml.) and show 
from it why the other three indicators would not be as satisfactory. Show 
clearly the positions of pH 4, 7, and 10 on the graph. 

691. Propionic acid is a monobasic acid with an ionization constant of 
1.6 X 10~ 5 . If 0.100 mole of the pure acid is dissolved in 100 ml. and titrated 
with 4.00 N NaOH, calculate the pH value (a) of the original solution, (b) of 
the solution when the acid is two-thirds neutralized, (c) at the equivalence 
point. 

84. Titration of Sodium Carbonate. In Fig. 3 (Sec. 82) curve 
(G) (<7) represents the titration of a solution of sodium carbonate 
with a half-normal solution of hydrochloric acid. It will be noted 
that there are two points of inflection. The first is at about 
pH = 9 and corresponds to the completion of the reaction 

CO,- + H+ -* HCO 3 - 

The second is at about pH = 4 and corresponds to the comple- 
tion of the reaction 

HCO 3 - + H+ -* C0 2 + H 2 

Phenolphthalein should therefore indicate the conversion of sodium 
carbonate to bicarbonate, and methyl orange should change color 
only when complete neutralization has taken place. Use is made 
of this principle in titrations of certain mixtures of substances as 
illustrated in the following section. 

85. Analyses Involving the Use of Two Indicators. The fact 
that certain indicators change color at different stages of a neu- 
tralization is sometimes made use of in volumetric work to deter- 
mine the proportions of the components of certain mixtures by 



200 CALCULATIONS OF ANALYTICAL CHEMISTRY 

the employment of two end points in a single titration. This 
may be brought about by means of two indicators, and the vol- 
umes of titrating solution required for the respective end points 
give a direct measure of the amounts of substances present. Only 
the two common indicators, methyl orange and phenolphthalein, 
will be considered. 

Assume a solution to contain only sodium hydroxide and inert 
impurities. The weight of NaOH present may be found by direct 
titration with a standard solution of any strong acid and with 
either methyl orange or phenolphthalein as the indicator. In 
either case, the color change will take place only when the alkali 
is completely neutralized, and the volume of standard acid used 
in the titration is a direct measure of the weight of NaOH present. 

If a solution contains only sodium carbonate and inert impurities 
and is titrated with standard acid, methyl orange being used as 
the indicator, the color change takes place only when the Na 2 CO 3 
has been completely neutralized. 

C0 8 - + H+ -* HCOr 
HC0 3 - + H+ - II 2 + C0 2 

The volume of acid required is a measure of the total alkaline 
strength of the sample and of the actual weight of Na 2 CO 3 present. 
In calculating, the equivalent weight of the Na 2 CO 3 would be 
taken as one-half of the molecular weight. On the other hand, 
if phenolphthalein were used as the indicator and the titration 
were carried out in the cold, the color change from pink to colorless 
would occur when the carbonate had been changed to bicarbonate. 

C0 8 - + H+ -> HC0 3 - 

The volume of standard acid required to titrate sodium carbon- 
ate to an end point with methyl orange as the indicator is twice 
that required if phenolphthalein is used as the indicator, since 
twice the number of hydrogen-ion equivalents is involved. The 
equivalent weight of the Na 2 C0 3 is identical in the latter case 
with the whole molecular weight Na 2 CO 3 /l, and the calculated 
weight of Na 2 CO 3 present is the same in the two cases. It is 
important to note that if, with phenolphthalein as the indicator, 
an excess of standard acid is added to the carbonate solution and 
the carbon dioxide is expelled by boiling ) the sodium carbonate will 



NEUTRALIZATION METHODS 201 

be completely neutralized. Neutralization of the excess acid with 
standard alkali will give a net volume of acid which will be the 
same as that used with methyl orange as the indicator. 

If a solution contains sodium bicarbonate and inactive impuri- 
ties, the NaHCOs may be titrated with standard acid, methyl 
orange being used as the indicator, or in boiling solution with 
phenolphthalein, in the latter case by adding excess acid and 
titrating back with alkali. 

HC(V + 11+ -> II 2 + C0 2 

The equivalent weight of NaHCO 3 in either case is identical to 
the molecular weight. As stated above, a cold solution of pure 
sodium bicarbonate gives no color with phenolphthalein and there- 
fore cannot be titrated with phenolphthalein as the indicator. 

There now remains the question of possible mixtures of the 
three alkalies just discussed. Altogether, there are the following 
theoretical possibilities : 

(a) NaOH 

(6) Na 2 C0 3 

(c) NallCOs 

(d) NaOH + Na 2 C0 3 

(e) Na 2 C0 3 + NaH(X) 3 
(/) NaOH + NaHCOa 

(0) NaOH + Na 2 C0 3 + NaHCO 3 

Inert impurities may be present in each case. The last two mix- 
tures, however, cannot exist in solution, for sodium hydroxide and 
sodium bicarbonate interact mole for mole to form the normal 
carbonate 

OH- + HCO 3 - -> C0 3 - + II 2 () 



Strictly speaking, these last two mixtures can exist when in the 
perfectly dry form, although this condition would be difficult to 
realize in practice. When they are treated with water the reaction 
takes place, forming the carbonate and leaving a possible excess 
of hydroxide or bicarbonate as the case may be. 

The mixtures ordinarily encountered in practice are those of 
(d) and (e) y viz., sodium hydroxide w r ith sodium carbonate and 
sodium carbonate with sodium bicarbonate. Two end points being 
used, it is possible to determine the proportions of the components 
of either of these mixtures even when inactive impurities are present. 



202 CALCULATIONS OF ANALYTICAL CHEMISTRY 

EXAMPLE I. MIXTURE OF HYDROXIDE AND CARBONATE. A 
1.200-gram sample of a mixture of sodium hydroxide and sodium 
carbonate containing inert impurities is dissolved and titrated 
cold with half-normal hydrochloric acid solution. With phenol- 
phthalein as an indicator, the solution turns colorless after the 
addition of 30.00 ml. of the acid. Methyl orange is then added, 
and 5.00 ml. more of the acid are required before this indicator 
changes color. What is the percentage of NaOH and of Na 2 CO 3 
in the sample? 

SOLUTION: If the acid is added slowly, the stronger base 
(NaOH) is neutralized first, as follows: 

OH- + H+ -+ H 2 O 

After this reaction is complete, the carbonate is converted to bi- 
carbonate. 

CCV* + H+ -> HCO 8 - 

At this point, the phenolphthalein changes from pink to colorless 
and a total of 30.00 ml. of acid has been added. Then the bicar- 
bonate formed is neutralized by 5.00 ml. more of acid. 

HC0 8 - + H+ -> H 2 + C0 2 

Since each mole of Na 2 CO 3 reacts with 1 mole of HOI to give 
1 mole of NaIIOO 3 and this, in turn, is neutralized by 1 mole 
of HOI, it follows that the volume of acid required to convert 
the Na 2 OO 3 into NaHOO 3 is the same as the volume required 
to neutralize the NaHCO 3 , viz., 5.00 ml. 

Therefore, the volume of acid necessary to neutralize completely 
the Na 2 C0 3 is 10.00 ml. Since the total volume added was 35.00 
ml., it is evident that 35.00 10.00 = 25.00 ml. were necessary 
to neutralize the NaOH. Hence, 

Percentage of NaOH = 



25.00 X 0.5000 X 

X 100 = 41.68 per cent 

A . J\J\J 

x A v} 9 

Percentage of NazCOs = 

10.00 X 0.5000 X 

X 100 = 22.08 per cent 



NEUTRALIZATION METHODS 



203 



These volume relationships are shown diagrammatically in 
Fig. 4. 

EXAMPLE II. MIXTURE OF CARBONATE AND BICARBONATE. 
A 1.200-gram sample of an impure mixture of sodium carbonate 
and sodium bicarbonate containing only inert impurities is dis- 
solved and titrated cold with half-normal hydrochloric acid solu- 



35 



30 



ml. 



25 



ml. 



ml. 



Phenolphthalein 



i 



NaCl 



Na 2 C0 3 



5 mi. 



i 



[ NaHCO 3 [< ~ changes color 

Methyl orange added 



I NaC1 U... Methyl orange 
I changes color 



5 ml. 



FIG. 4. 

tion. With phenolphthalein as an indicator, the solution turns 
colorless after the addition of 15.00 ml. of the acid. Methyl 
orange is then added, and 22.00 ml. more of the acid are required 
to change the color of this indicator. What is the percentage of 
Na 2 CO 3 and of NaHCO 3 in the sample? 

SOLUTION: When the acid is added slowly, the Na 2 COs is con- 
verted into NaHCO 3 . At this point, the phenolphthalein changes 
color, and 15.00 ml. of HC1 have been added. As in Example I, 
the same volume of HC1 as was used for the conversion of the 
Na 2 CO 3 into NaHCO 3 would be required to convert this NaIICO 3 
formed from the Na 2 C0 3 into NaCl, H 2 O, and C0 2 . It follows 
that 15.00 + 15.00 = 30.00 ml. of acid were required to neutralize 
completely the Na 2 CO 3 present in the sample. The total volume 
being 15.00 + 22.00 = 37.00 ml, it is evident that 37.00 - 30.00 = 



204 CALCULATIONS OF ANALYTICAL CHEMISTRY 

7.00 ml. of HC1 were required to neutralize the NaHCOs present 
in the original sample. Hence, 

Percentage of Na2CO3 = 

30.00 X 0.5000 X 



2,000 



1.200 
Percentage of NaHCO 3 = 

7.00 X 0.5000 X 



X 100 = 66.25 per cent 



NaHCO 3 
1,000 



1.200 



X 100 = 24.50 per cent 



Ans. 



These volume relationships are shown diagrammatically in 
Fig. 5. 

Na 2 CO 3 |<- Phenolphthalein added 



37 



15 



ml. 



ml. 



22ml. 



15 



ml. 



_ 





5 


NaHC0 3 1 




H 


NaCl 



Phenolphthalein changes color 
Methyl orange added 




Methyl orange 

changes color 



FIG. 5. 



86. Relation of Titration Volumes to Composition of Sample. 

In an analysis of the type discussed in this section, it is not al- 
ways true that the analyst is previously aware of the exact com- 
position of the sample. He may not know whether the sample 
contains hydroxide, carbonate, bicarbonate, or possible combina- 
tions of these components, and a qualitative test is not always 
conclusive. By means of a simple titration, however, and the 



NEUTRALIZATION METHODS 



205 



use of a double indicator or double end point, the composition 
of the alkali can be determined so far as these negative radicals 
are concerned. 

In this connection, let A represent the volume of standard acid 
required to titrate the cold solution to a change of color of phenol- 
phthalein, and let B represent the additional volume of the acid 
to continue the titration to a change of color of methyl orange. 
The following relationships exist: 



Active ions 
present 


Volume for 
first end point 


Additional volume 
for second end point 


OH- 


A 





HCOr 





B 


COr 


A 


B = A 


OOr + OH- 


A 


B < A 


cxv + H(XV 


A 


R> A 



Problems 

692. A sample of sodium carbonate containing sodium hydroxide and only 
inert impurities weighs 1.197 grains. It is dissolved and titrated in the cold 
with phenolphthalein as the indicator. The solution turns colorless when 
48.16 ml. of 0.2976 N HOI have been added. Methyl orange is added, and 
24.08 ml. more of the acid are required for complete neutralization. Calculate 
the percentages of NaOl I and Na 2 CO 2 in the sample. 

Ans. NaOH = 23.95 per cent, Na 2 CO 3 = 63.46 per cent. 

693. From the following data, calculate the percentages of Na 2 CO 3 and 
NaHCO 3 in a mixture in which they are the only alkaline components. 
Sample = 1 .272 grains. Volume of 0.2400 N IIC1 required for phenolphthalein 
end point = 26.92 ml. After addition of an excess of the standard acid arid 
the boiling out of the CO 2 , net additional volume of the acid required for the 
phenolphthalein end point = 50.21 ml. 

Ans. Na 2 OO 3 = 53.84 per cent, NaHCO 3 = 36.93 per cent. 

694. A sample of material contains for its active components NaOH, 
Na2CO 3 , NaIICO 3 , or possible mixtures of these. Two samples, each weighing 
1 .000 gram, are dissolved in water. To one sample phenolphthalein is added 
and the solution is titrated cold with 1.038 N acid, of which 17.96 ml. are re- 
quired. The other sample is titrated cold with methyl orange as an indicator, 
and 21.17 ml. of the same acid are required. What alkalies are present? 
Calculate the percentage of each. 

Ans. NaOH = 61.28 per cent, Na 2 CO 3 = 35.31 per cent. 

696. A chemist received different mixtures for analysis with the statement 
that they contained either NaOH, NaHCO 3 , Na 2 CO 3 , or possible mixtures 



206 CALCULATIONS OF ANALYTICAL CHEMISTRY 

of these substances with inert material. From the data given, identify 
the respective materials and calculate the percentage of each component. 
One-gram samples and 0.2500 normal HC1 were used in all cases. 

Sample I. With phenolphthalein as an indicator, 24.32 ml. were used. 
A duplicate sample required 48.64 ml. with methyl orange as an indicator. 

Sample II. The addition of phenolphthalein caused no color change. 
With methyl orange, 38.47 ml. of the acid were required. 

Sample III. To cause a color change in the cold with phenolphthalein 
15.29 ml. of the acid were necessary, and an additional 33.19 ml. were required 
for complete neutralization. 

Sample IV. The sample was titrated with acid until the pink of phenol- 
phthalein disappeared; this process required 39.96 ml. On adding an excess 
of the acid, boiling, and titrating back with alkali, it was found that the alkali 
was exactly equivalent to the excess acid added. 

Ans. I. 64.45 per cent Na 2 CO 3 . II. 80.79 per cent NaHCO 3 . III. 40.52 
per cent Na 2 CO 3 , 37.60 per cent NaHCO 3 . IV. 39.97 per cent NaOH. 

696. A sample is known to contain either NaOH or NaHC0 3 or Na 2 CO 3 or 
possible mixtures of these, together with inert matter. A 1.200-gram sample 
requires 42.20 ml. of N/2 HC1, using methyl orange as indicator. The same 
weight of sample requires 36.30 ml. of the acid using phenolphthalein indicator. 
Calculate the percentage of inert matter in the sample. 

Ans. 23.27 per cent. 

697. Pure dry NaOH and pure dry NaHCO 3 are mixed in the respective 
proportion by weight of 2:1, and the mixture is dissolved in water. Calcu- 
late to three significant figures the ratio of the volume of standard acid 
required with phenolphthalein as an indicator to the additional volume re- 
quired with methyl orange. 

Ans. 4.20. 

598. A mixture that contains KOH and K 2 CO 3 weighs a grams and, in 
the cold solution with phenolphthalein, requires 6 ml. of c normal acid. 
After methyl orange is added, d ml. of the acid are required. Calculate the 
percentage of KOH and of K 2 CO 3 . Reduce to simplest terms. 

A r> I-^TT 5.611(6 - d)c 
Ans. Per cent KOH = 

Per cent K 2 CO 3 = - 

a 

699. Solve the preceding problem with respect to a mixture of Na2CO 3 and 
NallCOs. Reduce to simplest terms. 

10.60 be 



Ans. Per cent Na 2 CO 3 



Per cent NaHC0 3 = 



a 
8.401 (d - 6)c 



a 

600. A liter of an alkali solution is prepared from 38.00 grams of pure 
NaOH and 2.00 grams of pure Na^COs. What is the true normality of the 



NEUTRALIZATION METHODS 207 

solution if completely neutralized? If this solution were used in a titration 
in the cold with phenolphthalein as an indicator, what normality should be 
taken for the alkali? 

Ans. 0.9874 N. 0.9686 N. 



601. Calculate the grams of NaOH and the grams of Na 2 CO 3 present in 
a mixture that on analysis gives the following data: Sample = 10.00 grams. 
The sample is dissolved in water, the solution is diluted to 250.0 ml., and 
25.00 ml. are taken for analysis. An end point with phenolphthalein in cold 
solution is obtained with 44.52 ml. of 0.5000 N HCL A new portion of the 
same volume requires 46.53 ml. of 0.5000 N HC1 for an end point with methyl 
orange. 

602. A mixture of soda ash and caustic soda weighs 0.7500 gram. It is 
dissolved in water, phenolphthalein is added, and the mixture is titrated 
cold with 0.5000 N HC1; the color disappears when 21.00 ml. have been added. 
Methyl orange is then added, and the titration continued until the pink color 
appears. This requires 5.00 ml. of acid in addition. Calculate the percentage 
of NaOH and of Na 2 CO 3 in the sample. 

603. A sample is known to contain NaOH or NaIICO 3 or Na2CO 3 or possible 
mixtures of these, together with inert matter. Using methyl orange, a 1.100- 
gram sample requires 31.40 ml. of IIC1 (1.00 ml. - 0.01400 gram CaO). 
Using phenolphthalein, the same weight of sample requires 13.30 ml. of the 
acid. Calculate the percentage of inert matter in the sample. 

604. The qualitative analysis of a powder shows the presence of sodium, a 
carbonate, and a little chloride. Titrated with methyl orange as an indicator, 
0.8000 gram of the powder reacts with 25.10 ml. of half-normal HC1, and the 
same weight reacts with 18.45 ml. of the acid with phenolphthalein in the cold 
solution. Compute the percentage composition of the original powder. 

605. A substance reacts alkaline in aqueous solution, and the alkalinity is 
due either to K 2 CO 3 and KOII or to K 2 C0 3 and KIIC0 3 . Compute the per- 
centage of each alkaline constituent from the following data: 

When phenolphthalein is the indicator (in a cold solution), 1.160 grams of 
powder react with 26.27 ml. of 0.3333 N HC1, and with 59.17 ml. when 
methyl orange is the indicator. Sample = 1.5000 grams. 

606. A solution of alkali is prepared from NaOH contaminated with 
Na 2 CO 3 . With phenolphthalein in the cold, 36.42 ml. of the alkali are re- 
quired to neutralize 50.00 ml. of 0.5280 N r H 2 SO 4 . With methyl orange as 
the indicator, 35.60 ml. of the alkali are required for the same amount of acid. 
How many grams of NaOH and of Na 2 CO 3 are contained in each milliliter 
of the alkali solution? 

87. Analysis of Phosphate Mixtures. Phosphoric acid ionizes 
in three steps. The ionization constant for the first hydrogen is 
1.1 X 10~ 2 ; for the second, 2.0 X 1Q- 7 ; and for the third, 3.6 X 10~ 13 . 



208 



CALCULATIONS OF ANALYTICAL CHEMISTRY 



H 3 P0 4 



In the titration of phosphoric acid with an alkali like NaOH, the 
replacement of the first hydrogen results in the formation of 
NaH 2 P0 4 (H 3 PO 4 + OH- -> H 2 PO 4 ~ + II 2 O). At approximately 
this point, methyl orange changes color. The replacement of the 
second hydrogen results in the formation of Na 2 HPO 4 (H 2 P0 4 ~ + 
OH~ >HPO 4 = + H 2 0). At approximately this point, phenol- 
phthalein changes color. The reverse titration of Na 3 PO 4 with a 
strong acid like HC1 results first in the formation of HPO^, at 

which point phenolphthalein 
changes color, and then in the 
formation of H 2 PO 4 ~, at which 
point methyl orange changes 
color. A titration of this sort is 
shown in Fig. 6. 

Only adjacent substances 
shown on the diagram can exist 
together in solution. Other com- 
binations interact. As in the 
case of the carbonate titrations 
of the preceding section, it is 
possible to analyze certain mix- 
tures of phosphates by means 
of titrations involving the use of 
two indicators. Actually, the 





J 


< 













. 


SC 



NaH 2 P0 4 



X 


> 


^ 







Q 


oo 















Na 2 HP0 4 


t I 


I 


/ 


^ 

g 



h ... Methyl orange 
changes color 



__ Phenolphthalein 
changes color 



Na 3 PO 4 



FIG. 6. 



titrations should be carried out 
on fairly concentrated solutions 
and at a temperature of about 55C. 

EXAMPLE. A sample, which is known to contain either Na 3 PO 4 , 
NaH 2 P() 4 , Na 2 HPO 4 , or possible mixtures of these, together with 
inert impurity, weighs 2.00 grams. When this sample is titrated 
with 0.500 N HC1, methyl orange being used, 32.0 ml. of the 
acid are required. The same weight of sample when titrated with 
0.500 N HC1, phenolphthalein being used, requires 12.0 ml. of 
the acid. What is the percentage composition of the sample? 

SOLUTION: From simple inspection of the diagram and con- 
sideration of the two volumes involved, it is evident that both 
SaaP0 4 and Na 2 HPO 4 are present. A volume of 12.0 ml. must 
lave been required to convert the Na 3 PO 4 to Na 2 HPO 4 ; and, since 
12.0 ml. more would be required to convert the Na 2 HPO 4 formed 



NEUTRALIZATION METHODS 209 

to NaH 2 PO 4 , a volume of 32.0 - (2 X 12.0) = 8.0 ml. is required 
to convert the original Na 2 HPO 4 to NaH 2 PO 4 . 



12.0 X 0.500 X 

X 100 = 49.2 per cent Na 3 P0 4 

Ans. 
8.0 X 0.500 X *'" 141J<1 



2.00 



Problems 



607. A sample that contains Na 3 PO4.12II 2 O, or Na 2 HPO 4 . 1 2II 2 O, or 
NaH 2 PO 4 .Il2O, or possible mixtures of these weighs 3.00 grams. When it is 
titrated with 0.500 N HOI, methyl orange being used, 14.0 ml. of the acid are 
required. A similar sample requires 5.00 ml. of 0.600 N NaOH, with phenol- 
phthalein. What is the percentage composition of the sample? 

Ans. 13.8 per cent NaH 2 PO 4 .H 2 O, 83.6 per cent Na 2 TTPO 4 .12H 2 O. 

608. A certain solution is known to contain any possible combinations 
of the following substances: HC1, Na 2 ITPO 4 , NaTI 2 PO 4 , H 3 PO 4 , NaOII. 
Titration of a sample with 0.500 N NaOH, phenolphthalein being used, re- 
quires 27.0 ml. of the base. With the same weight of sample arid methyl 
orange indicator, 17.2 ml. of the 0.500 N NaOII are required to obtain a color 
change. What components are present, and how many grams of each are 
present in the sample taken? 

Ans. 0.135 gram IIC1, 0.481 gram H 3 PO 4 . 

609. A certain solution is known to contain either IIC1 + H 3 PO 4 , or 
IT 3 PO 4 + NalI 2 PO 4 , or the three compounds existing alone. A sample is 
titrated with NaOH, requiring A ml. with methyl orange indicator; but the 
same weight of sample requires B ml. of the NaOH with phenolphthalein 
indicator. What relationship would exist between A and B to indicate the 
first combination, and what relationship would indicate the second com- 
bination? What relationship would indicate the presence of H 3 PO 4 alone? 

Ans. B > A, but B < 2A; B > 2A. B = 2A. 



610. A solution known to contain H 3 PO 4 , Na 2 HPO 4 , or NaH 2 PO 4 , or 
possible mixtures of these, weighs 1.10 grains. When it is titrated with 
0.520 N NaOH, 27.0 ml. are required to change the color of phenolphthalein, 
but only 10.0 ml. to change the color of methyl orange. What is the per- 
centage composition of the solution? 

611. A certain solution is known to contain any possible combinations of 
the following substances: HC1, Na 2 HPO 4 , NaII 2 PO 4 , H 3 PO 4 , NaOH, Na 3 PO 4 . 
Titration of a sample with 0.510 N HC1, methyl orange being used, requires 
28.1 ml. of the acid. With the same weight of sample and phenolphthalein 



210 CALCULATIONS OF ANALYTICAL CHEMISTRY 

indicator, 17.1 ml. of the HC1 are required. What components are present, 
and how many grams of each are in the sample taken? 

612. A series of solutions are known to contain H 3 PO 4 , NaII 2 PO 4 , Na2HP(>4, 
alone or mixed in varying proportions. In each case, the titration is made 
with 1.000 N NaOH to a pink color with phenolphthalein and the solution is 
then back-titrated with 1.000 N HC1 to a pink color with methyl orange. 
In each of the following four cases, determine which components are present 
and the number of millimoles of each: 

a. Initial titration 48.36 ml; back-titration 33.72 ml. 

6. Initial titration 37.33 ml.; back-titration 39.42 ml. 

c. Initial titration 24.36 ml.; back-titration 24.36 ml. 

d. Initial titration 36.24 ml.; back-titration 18.12 ml. 



CHAPTER XTII 

OXIDATION AND REDUCTION (REDOX) METHODS 
(OXIDIMETRY AND REDUCTIMETRY) 

88. Fundamental Principles. This phase of volumetric analy- 
sis has to do with the titration of an oxidizing agent with a standard 
solution of a reducing agent or the titration of a reducing agent 
with a standard solution of an oxidizing agent. This type of de- 
termination embraces the greater part of volumetric analysis, for 
the number of substances capable of oxidation or reduction is 
comparatively large. 

Oxidation is the increase in the positive direction of the elec- 
trical valence or oxidation number of an element or radical; re- 
duction is the decrease in electrical valence or oxidation number 
of an element or radical. Oxidation arid reduction must evidently 
take place simultaneously, for in any reaction of this type the 
oxidizing agent is always reduced and the reducing agent is always 
oxidized, and to the same degree. The methods of expressing 
concentration and the definitions given in Chap. IV hold true for 
solutions of oxidizing and reducing agents. Therefore, the rela- 
tionships existing between these agents are the same as those 
existing between acids and bases. However, it is necessary in the 
case of concentrations of solutions expressed in terms of normality 
to consider the hydrogen equivalent from a slightly different point 
of view. 

89. Equivalent Weights of Oxidizing and Reducing Agents. 
As in acidimetry and alkalimetry, the concentration of a solution 
of an oxidizing or reducing agent is best expressed in terms of its 
relation to the normal solution, and the gram-atom of hydrogen 
is taken as the ultimate unit. We must, however, consider the 
unit from the point of view of oxidation and reduction, thus: 

H <=> H+ + e 

Hydrogen ion is an oxidizing agent and is capable of being reduced 
to hydrogen gas (e.g., Zn + 2H+ Zn++ + H 2 ). Free hydrogen 

211 



212 CALCULATIONS OF ANALYTICAL CHEMISTRY 

is a reducing agent and is capable of being oxidized to hydrogen 
ion (e.g., 2Fe+++ + H 2 - 2Fe++ + 2H+). 

The conversion of one atom of hydrogen to the ion, or vice 
versa, involves a change of 1 in oxidation number and a transfer 
of one electron. To find the equivalent weight of an oxidizing or 
reducing agent we must, therefore, take that fraction of its formula 
weight so that in the oxidation or reduction process there will be 
involved the equivalent of a transfer of one electron. This will 
be accomplished (1) by dividing the formula weight of the sub- 
stance by the total change in oxidation number involved in the 
oxidation-reduction process, or (2) by dividing the formula weight 
of the substance by the number of electrons transferred per for- 
mula weight of substance. The gram-equivalent weight of an 
oxidizing agent is the equivalent weight in grams and is equivalent 
in oxidizing power to 1.008 grams of hydrogen as hydrogen ion. 
It is likewise equivalent in oxidizing power to 8.000 grams of 
oxygen. The gram-equivalent weight of a reducing agent is 
equivalent in reducing power to 1.008 grams of elementary hy- 
drogen gas. As will be seen later, a substance may have two 
different equivalent weights depending on whether it is used as 
an acid or as an oxidizing or reducing agent. As in acidimetry, 
a normal solution of an oxidizing or reducing agent contains one 
gram-equivalent weight of substance per liter of solution, or one 
gram-milliequivalent per milliliter. Hence, as in acidimetry, 

ml. X N = number of gram-milliequivalents 
and 

ml. 8 X N, X c x = grams* 

EXAMPLE I. How many grams of the following reducing 
substances constitute the gram-equivalent weight in each case: 
(a) FeS0 4 .7II 2 0, (6) SnCl 2 , (c) H 2 C 2 O 4 .2II 2 (oxalic acid), (d) 
KHC 2 4 .H 2 O (potassium binoxalate), (e) KHC 2 4 .H 2 C 2 4 .2H 2 O 
(potassium tetroxalate), (/) H 2 S (oxidized to S), (g) H 2 S (oxidized 
to H 2 S0 4 ), (K) Na 2 S 2 O 3 .5H 2 (oxidized to Na 2 S 4 6 ), (z) H 2 O 2 ? 

SOLUTION: (a) FeSO 4 .7H 2 O. In solution, this gives ferrous 
ions which can be oxidized to ferric ions. 



Fe++ -> 
+ Br 2 -> 2Fe+++ + 2Br~ 



OXIDATION AND REDUCTION METHODS 213 

Each ferrous ion changes in oxidation number by one unit and 
hence is equivalent in reducing power to the hydrogen unit. The 
molecular weight of FeS0 4 .7H 2 is therefore the equivalent weight 
as a reducing agent, and, expressed in grams, is equivalent in 
reducing power to 1.008 grams of hydrogen. 

FeS0 4 .7H 2 O 



1 



= 278.0 grams. Ans. 



(b) SnCl 2 . In solution, this gives stannous ions which can be 
oxidized to stannic ions: Sn ++ > Sn 4 " 1 "^ + 2e. The change in 
oxidation number is 2. The gram-molecular weight of SnCl2 is, 
therefore, equivalent in reducing power to 2 gram atoms of hy- 
drogen, or one-half the molecular weight represents the equivalent 
weight. 

? = 77.08 grams. Ans. 

z 

(c) H 2 C204.2H2(). In solution, this gives oxalate ions, C 2 O4 = r 
which can be oxidized to CO 2 gas. 

( 1 2 < V - - 2CO 2 + 2e 
e.g., 

5C 2 O 4 - + 2Mn0 4 ~ + 1GII+ 10CO, + 2Mn++ + 8IT 2 O 



The average oxidation number of carbon in the oxalate radical is 
+3 (actually one is +2, the other is +4). The oxidation number 
of carbon in CO 2 is +4. Each carbon, on an average, changes by 
one unit in oxidation number; but, since there are 2 carbon atoms 
in the oxalate radical, the average change for the oxalate radical 
is 2. The radical is, therefore, equivalent in reducing power to 
2 hydrogen atoms. 

H 2 C 2 4 .2H S ,. Q nQ A 
= 03.03 grams. Ans. 

z 

(d) KHCoC^.HaO. Here again, each molecule of the dissolved 
salt gives an oxalate ion which is oxidized to CO 2 as in the pre- 
ceding case. 

KHC 2 4 .H 2 O , 
~ = 73.07 grams. Ans. 

Z 

It should be noted that the equivalent weight of this salt as an 
acid is the molecular weight, or 146.14. Hence, a solution of po- 



214 CALCULATIONS OF ANALYTICAL CHEMISTRY 

tassium binoxalate which is 0.1 N as an acid is 0.2 N as a reducing 
agent. 

(e) KHC204.H 2 C204.2H 2 0. Since each molecule of this salt in 
solution gives 2 oxalate ions which are oxidized as above to C02, 
the equivalent weight of potassium tetroxalate as a reducing agent 

is 

KHC 2 04.H2C 2 4 .2H 2 PQ A 

- j - = 63.55 grams. Ans. 

When this salt is reacting as an acid, its equivalent weight is one- 
third of the molecular weight, or 84.73. A given solution of po- 
tassium tetroxalate has four-thirds the normality as a reducing 
agent that it has as an acid. 

(/) H 2 S. When this substance is oxidized to free sulfur, the 
change in oxidation number of sulfur is 2. 

= 17.04 grams. Ans. 

Zt 

(g) H 2 S. When this substance is oxidized to sulfate, the change 
in oxidation number of sulfur is from -2 to +6. 

H 2 S A 

~Q = 4.2M) grams. Ans. 
o 

(h) Na 2 S 2 O3.5H 2 O. In aqueous solution, this salt gives thio- 
sulfate ions which can be oxidized to tetrathionate ions. 



-> S 4 6 = + 2e 
e.g., 

2S 2 O 3 ~ + I 2 -> S 4 6 = + 21- 

In the thiosulfate radical, the average oxidation number of sulfur 
is +2; in tetrathionate, the average oxidation number of sulfur is 
+2J/2. The average change for each sulfur is %\ but, since in 
thiosulfate there are two sulfurs, the total change in oxidation 
number is 1. 

Na 2 S 2 3 .5H 2 01QO 

- ~ - = 248.2 grams. Ans. 

(i) H 2 2 . When hydrogen peroxide acts as a reducing agent, 
it is always oxidized to free oxygen; e.g., 

5H 2 O 2 + 2Mn0 4 - + 6H+ -> 5O 2 + 2Mn++ + 8H 2 O 



OXIDATION AND REDUCTION METHODS 215 

Average change in oxidation number of each oxygen atom in the 
hydrogen peroxide molecule is from 1 to 0. Total change for 
the molecule is 2. 



H 2 2 



= 17.01 grams. Ans. 



EXAMPLE II. How many grams of the following oxidizing sub- 
stances constitute the gram-milliequivalerit weight in each case: 
(a) K 3 Fe(CN) 6 , (6) KMn0 4 , (c) K 2 Cr 2 O 7 , (d) I 2 , (c) KBrO 3 (re- 
duced to bromide), (/) H 2 O 2 ? 

SOLUTION: (a) K 3 Fe(CN) 6 . In solution, this salt gives ferri- 
cyanide ions which are capable of being reduced to ferrocyanide 
ions. 



The change in oxidation number of the iron is from +3 to +2. 
K 3 Fe(CN) 6 n QOnQ A 

fobf) ^ - 3293 gram - ^ s - 

(ft) KMnOd. When reduced in the presence of acid, perman- 
ganate ions form manganous ions. 



MnOr + 8H f + 5e -> Mn++ + 4II 2 O 



MnOr + 5Fe++ + 8H+ -> Mn++ + 5Fe+++ + 4H 2 () 
Change in oxidation number of manganese is from +7 to +2. 
KMnOi 



5QQO 



r^oi^ A 

= 0.03161 gram. Ans. 



In alkaline solution, permanganate is reduced to MnO 2 with a 
change in oxidation number of 3 (from +7 to +4). 

MnO 4 ~ + 2H 2 O + 36 -> MnO 2 + 4011^ 

Here the equivalent weight is one-third of the molecular weight. 
(c) K 2 Gr 2 Or. Bichromate ions are ordinarily reduced to chromic 
ions. 



Cr 2 O 7 =s + 14H+ + 6e -> 2Cr^^- + 7II 2 O 



Cr 2 7 = + 6Fe^ + 14H+ -* 201^++ + 6Fc+++ + 7H 2 



216 CALCULATIONS OF ANALYTICAL CHEMISTRY 

The change in oxidation number of each chromium atom is from 
+6 to +3, or the change of the dichr ornate ion (since it contains 2 
chromiun ions) is 6. 

= 0.04903 gram. Ans. 

(d) I 2 . Iodine is reduced to iodide. 

I 2 + 2c -> 21- 

There is one unit change in oxidation number for each iodine 
atom, or two unit changes for the molecule. 

2 = 0.1209 gram. Ans. 



2,000 

0) KBrOa. Bromate reduced to bromide involves a change in 
oxidation number of the bromine from +5 to 1, or a change of 
6 units. 

Br0 3 - + 6H+ + 6e -> Bi- + 3H 2 O 

KBrOs 



6,000 



0.02784 gram. Ans. 



(/) II 2 02. As an oxidizing agent, hydrogen peroxide is reduced 
to water. 

H 2 O 2 + 2H+ + 2e - 2H 2 O 

Average change of each oxygen is from 1 to 2. Total change 
for the molecule is 2. 

= 0.01701 gram. Ans. 



90. Calculations of Oxidation and Reduction Processes. 

Since the concentration of solutions in oxidation and reduction 
titrations, like those in acidimetry and alkalimetry, is based on 
the hydrogen equivalent, the methods of calculation are identical. 
Thus, I liter of a normal solution of an oxidizing agent will exactly 
oxidize 1 liter of a normal solution of a reducing agent, or 2 liters 
of a half-normal solution. 

In titrating a reducing agent with a solution of oxidizing agent 
or an oxidizing agent with a solution of reducing agent, reasoning 



OXIDATION AND REDUCTION METHODS 217 

similar to that described in Sees. 73, 74, and 75 will evolve the* 
same general formulas as were there derived: viz., 

ml.. X N, X e x = grams* 
and 

ml., X N* X e x vx , x 

w . i , / V X 100 = per cent* 

Weight of sample 

The methods of solving the various types of problems described 
under Sees. 76, 77, and 78 likewise apply to oxidation and reduc- 
tion titrations. 

Problems 

613. Refer to Problem 35 and give the equivalent weights of the following 
oxidizing agents: (a) KsAsO 4 , (6) NaBrO 3 , (c) NaNO 2 , (d) NuaOfe. 

Ans. (a) 32.02, (6) 25.15, (c) 11.50, (d) 39.00. 

614. Refer to Probelm 34 and state what fraction of the molecular weight 
represents the milliequivalerit weight in the case of each of the following 
reducing agents: (a) Cr 2 (SO 4 ) 3 , (6) HC1, (c) H 2 S. 

Ans. (a) 1/6,000, (6) 1/1,000, (c) 1/2,000. 

616. Thirty milliliters of ferrous ammonium sulfate solution contain 1.176 
grams of pure FeSO 4 . (NH 4 ) 2 SO4.6II 2 O. Twenty milliliters of potassium 
dichromate solution contain 0.2940 gram of K 2 Cr 2 O 7 . Calculate (a) normality 
of the ferrous ammonium sulfate, (6) normality of the dichromate, (c) value 
of 1.000 ml. of ferrous solution in terms of the dichromate solution. 

Ans. (a) 0.1000 N, (fc) 0.2998 N, (c) 0.3335 ml. 

616. A solution of nitric acid is 3.00 N as an acid. How many milliliters' 
of water must be added to 50 ml. of the acid to make it 3.00 N as an oxidizing 
agent? Assume reduction of HNO 8 to NO. 

Ans. 100 ml. 

617. If 10.00 grams of E^Fe(CN) 6 .3H 2 O are dissolved in water and the 
volume made up to 500 ml., what is the normality of the solution as a reducing, 
agent? 

Ans. 0.04736 N. 

618. From the following data, calculate the ratio of the nitric acid as an 
oxidizing agent to the tetroxalate solution as a reducing agent (assume reduc- 
tion of NO 3 ~ to NO). 

1.000 ml. HNO 3 =0= 1.246 ml. NaOH 
1.000 ml. KHC 2 O 4 .H 2 C 2 O 4 .2H 2 O =0= 1.743 ml. NaOH 

Normality NaOH = 0.1200 
Ans. l.f 



619. To oxidize the iron in 1.00 gram of FeSO 4 .(NH 4 ) 2 SO 4 .6H 2 O requires. 
5.00 ml. of HNOs (3Fe++ + NO 3 - + 4H+ - 3Fe+++ + NO + H 2 0). How 



218 CALCULATIONS OF ANALYTICAL CHEMISTRY 

much water must be added to 500 ml. of this acid to make the concentration 
as an acid exactly one-tenth normal? 
Ans. 350 ml. 

620. A certain volume of a solution of KHC2O4.H2O would be oxidized in 
the presence of acid by an equal volume of 0.01000 molar KMnO 4 . How 
many milliliters of 0.01000 molar Ba(OH) 2 solution would be neutralized by 
20.00 ml. of the binoxalate? 

Ans. 25.00 ml. 

621. A method of standardizing KMnO 4 solution against a standard solu- 
tion of NaOH has been suggested. This consists in dissolving a small (un- 
weighed) amount of oxalic acid (or acid oxalate) in water and titrating with 
the standard alkali, using phenolphthalein indicator. The resulting solution 
is acidified with H 2 SO 4 and titrated with the KMnO 4 . If KHC 2 O 4 .H 2 O were 
used as the intermediate compound and the titrations required 10.58 ml. of 
0.2280 N NaOH and 38.10 ml. of KMnO 4 , calculate the normality of the 
KMriO 4 as an oxidizing agent. 

Ans. 0.1266 N. 

622. KIIC 2 O 4 .H2C2O 4 .2H 2 O and Na 2 C 2 O 4 are to be mixed in the proper 
proportion so that the normality of a solution of the mixture as a reducing 
agent will be 2.15 times the normality as an acid. Calculate the proportion. 

Ans. 1:0.644. 

623. Calculate the normality as an acid and as a reducing agent of a 
solution made by dissolving a mixture of 20.00 grams of H 2 C 2 O 4 .2H 2 O, 10.00 
grams of KHC 2 O 4 , and 15.00 grams of KHC20 4 .H 2 C2O 4 .2H 2 O in water and 
diluting to exactly 1 ,000 ml. 

Ans. 0.5728 N, 0.7096 N. 



624. Refer to Problem 37 and give the gram-milliequivalent weights of the 
following oxidizing agents: (a) BiO 2 , (&) K 3 Fe(CN) 6 . 

625. Refer to Problem 36 and state what fraction of the molecular weight 
represents the equivalent weight in the case of each of the following reducing 
agents: (a) KNO 2 , (6) UO 2 C1 2 , (c) MnSO 4 , (d) VOSO 4 . 

626. In the analysis of chrome iron ore, chromium is oxidized by fusion 
to chromate and determined by titration with ferrous ammonium sulfate. 
What is the equivalent weight in terms of (a) Cr 2 O 3 and (6) Cr? 

627. What volume of HC1 solution is theoretically required to dissolve 
1.000 gram of pure iron out of contact with the air, if 3.00 ml. of the acid will 
neutralize that volume of KOII solution which will react with 6.00 ml. of a 
potassium acid oxalate solution that is 2.00 N as a reducing agent? 

628. When 25.00 ml. of IIC1 are treated with AgNO 3 , a precipitate of AgCl 
weighing 0.5465 gram is obtained; 24.36 ml. of the IIC1 exactly react with 
27.22 ml. of NaOH solution; 26.24 ml. of the NaOH exactly react with 30.17 
ml. of KHC 2 O 4 .H 2 C2O 4 .2H 2 O solution. How much water must be added to a 
liter of the oxalatg solution to make it exactly 0.02500 N as a reducing agent? 



OXIDATION AND REDUCTION METHODS 219 

629. The hydrogen peroxide sold for medicinal purposes is often labeled 
" 10 volume" which means that on ordinary decomposition it yields ten times 
its volume of oxygen. What would be the normality of such a solution as 
an oxidizing agent? As a reducing agent? 

630. A certain volume of KHC 2 O4.H2C2O4.2H 2 O solution would be neu- 
tralized by an equal volume of 0.01000 molar Na2CO3 solution. How many 
milliliters of 0.02000 molar K 2 Cr 2 7 would be required to oxidize 25.00 ml. of 
the tetroxalate solution? 

91. Permanganate Process. Potassium permanganate is exten- 
sively used as an oxidimetric standard. It serves as its own in- 
dicator. A normal solution contains one-fifth the gram-molecular 
weight per liter (see Sec. 89, Example 116) if used in the presence 
of acid. A standard solution of potassium permanganate is used 
in three ways: 

1. It is used in the presence of acid in the direct titration of a 
large number of oxidizable cations and anions. Among them are 
the following: 

SUBSTANCE OXIDIZED TO 



vo++ vo 3 - 

C 2 or co 2 

NO 2 ~ NOr 

so 3 - sor 

H 2 O 2 O 2 

MO+++ MoO 



++ 



U++++ UO 2 

2. It is used in the presence of acid in the indirect titration 
of a large number of reducible substances. In each case a meas- 
ured amount of a reducing agent (e.g., a ferrous salt or an oxalate) 
is added, and, after reduction is complete, the excess reduc- 
ing agent is titrated with standard permanganate (see Sec. 76). 
Among the many substances that can be determined in this way 
are the following: 

StrBSTAN REDUCED TO 



Cr 2 O 7 - 

MnO 2 , Mn 3 O 4 Mn ++ 

PbO 2 . Pb 2 O 3 . Pb.,0, 



220 CALCULATIONS OF ANALYTICAL CHEMISTRY 

3. It is used in neutral or alkaline solution in the titration of a 
very few substances. In these cases the permanganate is reduced 
to MnC>2, which precipitates. The permanganate, therefore, has 
an oxidizing power only three-fifths of what it has when used in 
the presence of acid (see Sec. 89). This fact must be made use 
of in the calculations of such analyses (see Example VI below). 

SUBSTANCE OXIDIZED TO 

Mn ++ MnO 2 

HCOOH (formic acid) CO 2 

EXAMPLE I. What is the normality of a solution of potassium 
permanganate if 40.00 ml. will oxidize that weight of potassium 
tetroxalate, KHC 2 O4.H2C2O4.2H 2 0, which requires 30.00 ml. of 
0.5000 N sodium hydroxide solution for its neutralization? 

SOLUTION: The amount of tetroxalate that requires 30.00 ml. 
of 0.5000 N NaOH for neutralization is 

or* f\f\ vx f\ cr\r\r\ vx KHC2O4.H2C2O4.2H2O ., ^. 

30.00 X 0.5000 X s~000 ^ *-^ grams 

The normality of the permanganate is therefore 

1.271 



40 00 X 



= 0.5000. Ans. 



4,000 

The same result is more simply obtained by setting up the entire 
equation before performing any of the operations, when it will be 
found that the molecular weights of the potassium tetroxalate 
cancel and need not be calculated. Thus, the weight of potassium 
tetroxalate neutralized by 30.00 ml. of 0.5000 N NaOH is 

30.00 X 0.5000 X 

and the weight of potassium tetroxalate oxidized by 40.00 ml. of 
x normal KMnO 4 is 

in flfl y r y KHC 2 04.H2C 2 04.2H20 
40.00 XxX 



Since these two expressions are equal to each other, the equality 
may be expressed by an equation in which the molecular weights 
of the potassium tetroxalate cancel and x gives the value 0.5000 
for the normality of the permanganate. 



OXIDATION AND REDUCTION METHODS 221 

EXAMPLE II. What is the percentage of iron in a sample of 
iron ore weighing 0.7100 gram if, after solution and reduction of 
the iron with amalgamated zinc, 48.06 ml. of KMnO 4 (1.000 ml. =0= 
0.006700 gram Na 2 C 2 O 4 ) are required to oxidize the iron? How 
many grams of KMn0 4 are contained in each milliliter of the 
solution? 
SOLUTION : 



Normality of the KMn0 4 = = 0.1000 N 

JN a2L/2vJ 4 

~27KKT 

48.06 X 0.1000 X y^r 

- QJ^Q -^- X 100 = 37.79 per cent Fe. Ans. 

Each milliliter of normal KMn0 4 contains KMn0 4 /5,000 = 
0.03161 gram. 

Each milliliter of this KMnO 4 contains 0.03161 X 0.1000 = 
0.003161 gram. Ana. 

EXAMPLE III. How many grams of II 2 02 are contained in a 
solution that requires for titratioii 14.05 ml. of KMn0 4 of which 
1.000 ml. =0= 0.008378 gram Fe (i.e., will oxidize that amount of 
iron from the divalent to the trivalent state)? How many grams 
and how many milliliters of oxygen measured dry and under stand- 
ard conditions are evolved during the titration? 
SOLUTION : 



Normality KMn0 4 = /1 = 0.1500 N 

re/I, 



Grams H 2 O 2 = 14.05 X 0.1500 X ~~ = 0.03584 gram. Ans. 

Each mole of H 2 O 2 corresponds to a mole of O 2 evolved [see 
Sec. 89, Example !(*)]- Therefore, 

Grams 2 evolved = 14.05 X 0.1500 X x~c = 0.03372 gram. 

A,UUU 

Ans. 

Each mole of O 2 occupies 22,400 ml. 
Therefore, 

99 4OA 

Milliliters O 2 evolved = 14.05 X 0.1500 X ~ = 23.60 ml. 

A na. 



222 CALCULATIONS OF ANALYTICAL CHEMISTRY 

EXAMPLE IV. What is the percentage of MnO 2 in impure py- 
rolusite if a sample weighing 0.4000 gram is treated with 0.6000 
gram of pure H 2 C20 4 .2H 2 and dilute H 2 S0 4 and after reduction 
has taken place (MnO 2 + H 2 C 2 O 4 + 2H+ -> Mn++ + 2CO 2 +2H 2 O) 
the excess oxalic acid requires 26.26 ml. of 0.1000 N KMn0 4 for 
titration? 
SOLUTION: 



Milliequivalents of H 2 C 2 4 .2H 2 O used = = 9.526 



Milliequivalents of KMnO 4 used = 26.26 X 0.1000 = 2.626 
Net milliequivalents = 9.526 - 2.626 = 6.900 

6.900 X MnOz/2,000 ,__ _. __ A ,, ~ 

- 040QA - x 10 ~ 74 -97 per cent Mn0 2 . Ans. 

EXAMPLE V. What would be the milliequivalent weight of 
PbsO 4 and of Pb in the calculation of the analysis of red lead 
(impure PbsC^) by a method similar to that of the preceding ex- 
ample (Pb 3 4 + H 2 C 2 4 + 3S0 4 = + 6H+ - 3PbS0 4 + 2CO 2 + 
4H 2 0)? 

SOLUTION: The oxidation number of lead changes from an 
average of 2% (in Pb 3 O 4 ) to 2 (in PbSO 4 ). Each lead therefore 
changes by an average of % unit; 3 leads change by 2 units. 
Hence, 



Me. wt. Pb 3 4 = ~ = 0.3428. Ans. 

Z } \)(j(j 
qpu 

Me. wt. Pb = ~ = 0.3108. Ans. 



EXAMPLE VI. A steel containing 0.90 per cent Mn is analyzed 
by the three standard methods below, in each case with a 2.50- 
gram sample, 0.0833 N KMnO 4 and 0.100 N FeS0 4 solutions. 
Calculate in each case the volume of KMnO 4 required. 

SOLUTION : Bismuthate Method. The Mn is oxidized to KMnO 4 
and after reduction with 25.0 ml. of the standard FeSO 4 (MnO 4 ~ + 
5Fe++ + 8H+ -> Mn++ + 5Fe+++ + 4H 2 0) the excess ferrous iron 
is titrated with the standard KMnO 4 . 

Let x = milliliters of KMn0 4 used in the titration. 



OXIDATION AND REDUCTION METHODS 223 

Milliequivalents of FeS0 4 used = 25.0 X 0.100 = 2.50 
Milliequivalents of KMnO 4 used = x X 0.0833 

Net milliequivalents = 2.50 - 0.0833s 



x = 5.42 ml. Ans. 

Chlorate (Williams) Method. The Mn is oxidized with KClOs 
to MriO 2 which is filtered and dissolved in 25.0 ml. of the standard 
FeS0 4 (MnO 2 + 2Fe++ + 4H+^Mn^ + 2Fe+++ + 2H 2 0). The 
excess FeSO 4 is titrated with the standard KMn0 4 . 

Let x = milliliters of KMnO 4 used in the titration. 

Milliequivalents of FeSO 4 used = 25.0 X 0.100 = 2.50 
Milliequivalents of KMnO 4 used = x X 0.0833 

Net milliequivalents = 2.50 - 0.0833z 



x = 20.2 ml. Ans. 

Volhard Method. The Mn is titrated directly with KMnO 4 in 
a solution kept neutral with ZnO (3Mn++ + 2Mn0 4 ~ + 2ZnO -> 
5Mn0 2 + 2Zn++). 

Let x = milliliters of KMn0 4 used in the titration. 

Tn this case the normality of the KMnO 4 cannot be taken as 
0.0833 because it is used in neutral solution where the change 
in oxidation number of its manganese is 3 instead of 5. In other 
words, the oxidizing power of KMnO 4 in neutral solution is only 
three-fifths as great as it is in acid solution. In this particular 
case the normality is 0.0833 X %. The change in oxidation num- 
ber of the titrated Mn is 2. 

s(0.0833 X ?flMn/2,000 

- ___ -- x luu = u.yu 

x = 16.4 ml. Ans. 

EXAMPLE VII. A 1.00-gram sample of steel containing 0.90 
per cent Mn is analyzed by the persulfate method whereby the 
manganese is oxidized to permanganate by ammonium persulfate 
and the resulting permanganate is titrated with a standard solu- 
tion of sodium arsenite. If 7.68 ml. of arsenite solution (0.0400 



224 CALCULATIONS OF ANALYTICAL CHEMISTRY 

molar in NasAsOa) are required and the arsenite is oxidized to 
arsenate in the titration, to what average oxidation number was 
the manganese reduced in the titration? 
SOLUTION: 

0.0400 molar Na 3 AsO 3 = 0.0800 normal 

Let x = change in oxidation number of Mn during titration 



7.68 X 0.0800 X = 0.0090 

x X 1,000 

Solving, 

x = 3.75 

Oxidation number of Mn in reduced form = 7 3.75 = 3.25. Ans. 

EXAMPLE VIII. If 1.000 ml. of a solution of KMn0 4 is equiva- 
lent to 0.1000 millimole of NaCHO 2 (sodium formate) in the fol- 
lowing titration: 3CHO 2 ~ + 2MnO 4 ~ + H 2 O -> 3CO 2 + 2MnO 2 + 
5OH-, what is the value of the KMnO 4 in terms of grams of CaO 
in the volumetric method for calcium in which that element is 
precipitated as CaC 2 4 .H 2 O and the precipitate is filtered, dis- 
solved in dilute H 2 S0 4 , and the oxalate titrated with perman- 
ganate? 
SOLUTION: 

0.1000 millimole NaCHO 2 = 0.2000 milliequivalent 

(since in the titration the oxidation number of C changes from 
+2 to +4) 

Normality KMnO 4 = 0.2000 N 

This normality applies only to the above type of titration in 
which the oxidation number of Mn in KMnO 4 changes by 3 units. 
Therefore, 

Normality KMn0 4 (presence of acid) = 0.2000 X % = 0.3333 N 

Each atom of Ca is combined with and equivalent to 1 mole 
of oxalate. Since the milliequivalent weight of the oxalate radical 
is its molecular weight over 2,000, the milliequivalent weight of 
CaO must be its molecular weight over 2,000. 



1.000 X 0.3333 X ~ = 0.009347 gram CaO. Ans. 



OXIDATION AND REDUCTION METHODS 225 

92. Bichromate Process. Potassium dichromate is occasionally 
used as an oxidimetric standard. With chemical indicators (e.g., 
potassium ferricyanide as an external indicator, or diphenylamine 
sulfate as an internal indicator), the use of dichromate in direct 
titrations is restricted to the titration of ferrous iron. Oxidizing 
substances can be determined by the dichromate process, as in 
the permanganate process, by the addition of a measured excess 
of a ferrous salt and the titration of the excess with the standard 
solution. Potassium dichromate titrations have greater applica- 
bility in potentiometric titrations where chemical indicators are 
not necessary. 

The normal solution of potassium dichromate contains one-sixth 
of the gram-molecular weight of K 2 Cr 2 Or per liter (see Sec. 89). 

EXAMPLE I. What is the percentage of FesOs in a sample of 
limonite ore (impure Fe 2 C>3) if the iron from a 0.5000-gram sample 
is reduced and titrated with 35.15 ml. of a potassium dichromate 
solution of which 15.00 ml. are equivalent in oxidizing power 
to 25.00 ml. of a potassium permanganate solution which has an 
"iron value" of 0.004750 gram? (This last expression is a con- 
ventional means of signifying that 1.000 ml. of the solution will 
oxidize 0.004750 gram of iron from the divalent to the trivalent state.} 
SOLUTION: 



Normality of KMnO 4 = t OQO'V Fe/1 000 = - 08506 
Normality of K 2 Cr 2 7 = 0.08506 X 25.00/15.00 = 0.1418 

35.15 X 0.1418 x|^ 

' X 100 = 79.60 per cent Fe 2 3 . Ans. 

U.oUUU 

EXAMPLE II. Fusion with Na 2 2 oxidizes the chromium in a 
0.2000-gram sample of chromite ore to chromate. The addition 
of a 50-ml. pipetful of ferrous sulfate solution reduces this in acid 
solution to chromic salt (Cr 2 Or + 6Fe++ + 14H+ -> 2O+++ + 
GFe 4 " 4 " 4 " + 7H 2 O), and the excess ferrous ions are titrated with 
7.59 ml. of 0.1000 N K 2 O 2 O 7 . Each pipetful of ferrous solution 
is equivalent to 47.09 ml. of the standard K 2 Cr 2 7 solution. What 
is the percentage of Cr in the sample? What weight of sample 
of the chromite ore should be taken such that the milliliters of 



226 CALCULATIONS OF ANALYTICAL CHEMISTRY 



standard 0.1000 N K 2 Cr 2 07 that are equivalent to the ferrous 
solution added, minus the milliliters of K^CrOy used in the titra- 
tion, will equal the percentage of Cr 2 03 in the sample? 

SOLUTION : Net K 2 Cr 2 07 solution (equivalent to the Cr in the 
ore) = 47.09 - 7.59 = 39.50 ml. 

39.50X0. 1000 Xg~ 

nonnn '- X 100 = 34.23 per cent Cr. Ans. 

(j.ZOOO 

The second part of this problem merely states that the net 
volume of 0.1000 N K 2 Cr 2 O 7 (i.e., the milliliters equivalent to the 
Cr in the sample) is equal in value to the percentage of Cr 2 0a. 

aX0.1000xCr 2 O 3 /6,000 _ 

" /\ JLUU O> 

X 

x = 0.2533 gram. Ans. 

93. Ceric Sulfate or Cerate Process. Cerium in the valence of 
4 is a very powerful oxidizing agent, the yellow 4-valent eerie or 
complex cerate ions being reduced to colorless 3-valent cerous ions. 



or, 



A solution of eerie sulfate is satisfactory for oxidimetry titrations 
and has certain advantages over potassium permanganate, par- 
ticularly with respect to its greater stability and its lesser tendency 
to oxidize chloride ions. In the titration of reducing substances 
that in solution are colorless, the yellow color of the excess eerie 
ions serves as a fairly satisfactory indicator. Titration of ferrous 
ions can be accomplished with orthophenanthroline ("ferroin") 
as an internal indicator. The potential of the indicator in its 
two states of oxidation lies between those of ferrous-ferric iron 
and cerous-ceric cerium. 

Fe++ <=> Fe+++ + e 
Ferroin' (red) <= ferroin" (blue) + e 



Ce++++ + e - 

+ e _> CC+++ + 3S0 4 



Ceric sulfate is particularly satisfactory in potentiometric titra- 
tions. 



OXIDATION AND REDUCTION METHODS 227 

EXAMPLE. What weight of limonite should be taken so that 
after solution in HC1 and reduction of the iron, the volume of a 
standard eerie solution required for titration will be one-half the 
percentage of Fe 2 O 3 in the sample (6.00 ml. of the eerie solution = 
2.00 ml. KHC 2 O 4 solution - 3.00 ml. of 0.0800 N NaOH)? 
SOLUTION: 

KHC 2 O 4 soln. = 0.0800 X 3.00/2.00 
= 0.120 N os an acid 
= 0.240 N as a reducing agent 
Ceric soln. = 0.240 X 2.00/6.00 - 0.0800 N 

1 X 0.0800 X Fe 2 () 3 /2,000 ^ , nn n 
~~~~ ~~~ X 1UU & 

x = 0.320 gram. Ans. 

Problems 

631. A solution of permanganate contains 2.608 grams of KMnO 4 per 750 ml. 
What is the normality of the solution and what is the value of 1.000 ml. in 
terms of (a) Fe 2 O 3 , (6) Fe, (c) KHC 2 O 4 , (d) TI 2 O 2 , (e) U(SO 4 ) 2 (oxidized to UO 2 ++)? 

Ans. 0.1100 N. (a) 0.008784 gram, (6) 0.006144 gram, (r) 0.007046 gram, 
(d) 0.001871 gram, (c) 0.02367 gram. 

632. Given a solution of KMnO 4 of which 1.000 ml. =c= 1.000 ml. KHC 2 O 4 
solution =c= 1.000 ml. NaOH =0= 0.1000 millimole of KHC 8 H 4 O 4 (potassium 
acid phthalate). What is the value of 1 ml. of it in terms of grams of Fe 2 O 3 ? 
How many millimoles of Mn are present in each milliliter? 

Ans. 0.01597 gram. 0.04000 millimoles. 

633. How many grams of KMnO 4 are contained in a liter of potassium 
permanganate if a certain volume of it will oxidize a weight of potassium 
tetroxalate requiring one-half that volume of 0.2000 N potassium hydroxide 
solution for neutralization? 

Ans. 4.214 grams. 

634. What is the normality of a solution of potassium permanganate if 
50.13 ml. will oxidize that weight of KHC 2 O 4 which requires 43.42 ml. of 
0.3010 N sodium hydroxide for neutralization? 

Ans. 0.5214 N. 

636. 1.000 ml. KHC 2 4 .H 2 C2O4.2H 2 O =c= 0.2000 ml. KMnO 4 

1.000 ml. KMnO 4 - 0.1117 gram Fe 

What is the normality of the tetroxalate solution when used as an acid? 
Ans. 0.3000 N. 

636. Given two permanganate solutions. Solution A contains 0.01507 
gram of KMnO 4 per milliliter. Solution B is of such strength that 20.00 ml. ^ 



228 CALCULATIONS OF ANALYTICAL CHEMISTRY 

0.1200 gram Fe. In what proportion must the two solutions be mixed in 
order that the resulting solution shall have the same oxidizing power in the 
presence of acid as 0.3333 N K 2 Cr2O 7 has? 
A Vol. A * _ 7A 

Ans - v^Ts = L576 ' 

637. How many milliliters of K 2 Cr 2 O7 solution containing 25.00 grams of 
anhydrous salt per liter would react with 3.402 grams of FeSO 4 .7H 2 in dilute 
acid solution? 

Ans. 24.00 ml. 

638. If 25.0 ml. of ferrous sulfate solution in sulfuric acid require 31.25 
ml. of 0.100 N K 2 Cr 2 O 7 solution for oxidation, how much water must be added 
to 200 ml. of the reducing solution to make it exactly one-twentieth normal? 

Ans. 300 ml. 

639. How many grams of pure K 2 Cr 2 O 7 must be weighed out, dissolved, 
and diluted to exactly 700 ml. to make a solution which, when used in the 
titration of iron in a sample of ore, shall be of such a strength that four times 
the number of milliliters used with a half-gram sample will represent one-half 
the percentage of FeO in the sample? 

Ans. 19. 12 grams. 

640. How many grams of pure Pb 3 O 4 ( = PbO 2 .2PbO) must be dissolved in 
a mixture of 30 ml. of 6 N H 2 SO 4 and 2.000 millimoles of KHC 2 O 4 .II 2 C2O 4 .2H 2 O 
so that 30.00 ml. of 0.1000 N KMnO 4 will be required for the excess oxalate? 

Ans. 1.714 grams. 

641. What weight of spathic iron ore (impure FeCO 3 ) should be taken for 
analysis such that the number of milliliters of KMnO 4 (1.000 ml. =c= 0.3000 ml. 
of potassium tetroxalate solution which is one-fourth normal as an acid) used 
in titration will be twice the percentage of FeO in the ore? 

Ans. 1.438 grams. 

642. If 0.9000 gram of oxalic acid (H 2 C 2 O 4 .2H 2 O) is allowed to react with 
0.5000 gram of pyrolusite and the excess oxalic acid is titrated with per- 
manganate, what must be the normality of the permanganate in order that 
one-half the percentage of MnO 2 may be obtained by subtracting the buret 
reading from the volume A of the permanganate equivalent to the 0.9000 
gram of oxalic acid used? What is the value of A? 

Ans. 0.2300 N. 62.09ml. 

643. A sample of steel weighing 2.20 grams and containing 0.620 per cent 
of Mn is dissolved, and the manganese is eventually titrated in neutral 
solution with standard KMnO 4 . (3Mn++ + 2MnO 4 ~ + 2H 2 O -> 5MnO 2 + 
4H+.) If 6.88 ml. are required, what is the value of each milliliter of the 
KMnO 4 in terms of H 2 C 2 O 4 .2H a O? 

Ans. 0.00756 gram. 



OXIDATION AND REDUCTION METHODS 229 

644. Sodium formate, NaCHO2, can be titrated in neutral solution accord- 
ing to the equation: 3CHO-T + 2MnOr + H 2 O -> 2Mn0 2 + 3CO 2 + 50H~. 
If 10.00 ml. of the KMn0 4 are equivalent to 0.08161 gram of sodium formate 
by this method, (a) what is the "iron value " of each milliliter of the KMnO 4 
(6) what is the value of each milliliter in terms of rnillimoles of H 2 O 2 , (c) what 
is the value of each milliliter in terms of grams of CaO, and (d) what is the 
value of each milliliter in terms of grains of Mn by the Volhard method? 

Ans. (a) 0.02234 gram, (6) 0.2000 millimoles, (c) 0.01122 gram, 
(d) 0.006598 gram. 

646. Calcium can be precipitated as CaC 2 O 4 .H 2 O, and the precipitate 
filtered, washed, and dissolved in dilute II 2 S0 4 . The oxalic acid formed can 
then be titrated with potassium permanganate. If a 0.1000 N solution of 
KMnO 4 is used, calculate the value of 1.000 ml. in terms of (a) Ca, (&) CaO, 
(r) CaCO 3 . 

Ans. (a) 0.002004 gram, (6) 0.002804 gram, (c) 0.005004 gram. 

646. If the iron in a 0.1500-gram sample of iron ore is reduced and sub- 
sequently requires 15.03 ml. of permanganate for oxidation, what is the purity 
of the ore expressed as percentage of (a) Fe, (6) FeO, (c) Fe 2 O 3 ? (4.000 ml. 
KMnO 4 o 3.000 ml. KHC 2 O 4 .H 2 C,O 4 solution =c= 3.000 ml. 0.1000 N NaOII.) 

Ans. (a) 55.95 per cent, (6) 71.96 per cent, (c) 80.00 per cent. 

647. What is the percentage purity of a sample of impure H 2 C 2 O 4 .2II 2 O if a 
sample weighing 0.2003 gram requires 29.30 ml. of permanganate solution, of 
which 1.000 ml. - 0.006023 gram Fe? 

Ans. 99.53 per cent. 

648. To a half-gram sample of pyrolusite is added a certain weight of 
oxalic acid (II 2 C 2 O 4 .2II 2 O). After reaction in acid solution is complete, the 
excess oxalic acid requires 30.00 ml. of 0.1000 N KMnO 4 for oxidation. If the 
pyrolusite is calculated to contain 86.93 per cent MnO 2 , what is the weight ot 
oxalic acid added? (MnO 2 + H 2 C 2 O 4 + 211+ -> Mn++ + 2CO 2 + 2H 2 O.) 

A ns. 0.8194 gram. 

649. One hundred milliliters of K 2 Cr 2 O 7 solution (10.0 grams per liter), 
5.00 ml. of 6 N H 2 SO 4 , and 75.0 ml. of FeSO 4 solution (80.0 grams FeSO 4 .7H 2 O 
per liter) are mixed and the resulting solution is titrated with 0.2121 N 
KMnO 4 . Calculate the volume required. 

Ans. 5.63 ml. 

660. In analyzing a one-gram sample of hydrogen peroxide with permanga- 
nate, what must be the normality of the KMn0 4 in order that the buret reading 
shall represent directly the percentage of H 2 O2? 

Ans. 0.5880 N. 

661. A sample of magnetite (impure Fe 3 O 4 ) is fused with Na 2 O 2 and all the 
iron thus oxidized to the ferric state. After leaching with water andiacidifying, 
the total iron is determined by reduction in a Jones reductor and titration with 



230 CALCULATIONS OF ANALYTICAL CHEMISTRY 

standard KMnO 4 . Volume of KMnO 4 required = 30.10 ml. It is of such 
concentration that 2.000 ml. - 3.000 ml. KHC 2 O 4 solution o 2.000 ml. 
NaOH =0= 1.000 ml. H 2 SO 4 =0= 0.008138 gram ZnO. What is the normality of 
the KMn0 4 and how many grams of Fe 3 O 4 are present in the sample of mag- 
netite? 

Ans. 0.2000 N, 0.4646 gram. 

662. Six millimoles of MnO are ignited in air (6MnO + O 2 * 2MnsO 4 ) and 
the resulting Mn 3 O 4 (= MnO2.2MriO) is dissolved in a solution containing 
25 ml. of 6 N H 2 SO 4 and A grams of FeSO 4 .(NH 4 ) 2 SO 4 .6H 2 O. The manganese 
is reduced by the ferrous ions completely to the divalent form. If the excess 
ferrous ions require 12.00 ml. of KMnO 4 (containing 0.05000 millimole of 
KMnO 4 per ml.) calculate the value of A. 

Ans. 2.745 grams. 

663. A sample of steel weighs 2.00 grams and contains 0.55 per cent Mn. 
After dissolving in HNO 3 the manganese is oxidized to permanganate with 
solid BiO 2 and the excess BiO 2 is filtered off. Excess FeSO 4 .7H 2 O (dissolved 
in water) is now added and the excess ferrous ions require 20.0 ml. of 0.200 N 
KMnO 4 . How many grams of FeSO 4 .7H 2 O were used? If the reduction had 
been made with Na 2 C 2 O 4 instead of with FeSO 4 .7II 2 O, how many millimoles of 
Na 2 C 2 O 4 should have been added in order for 20.0 ml. of the KMnO 4 to be 
required for the excess oxalate? 

Ans. 1.39 grams. 2.50 millimoles. 

664. A sample of chromite contains 30.08 per cent Cr 2 O 3 . After fusion of a 
0.2000-gram sample with Na 2 O 2 and dissolving in acid, how many grams of 
FeSO 4 .(NH 4 ) 2 SO 4 .6II 2 O should be added so that the excess ferrous ions will 
require 15.00 ml. of 0.6011 N K 2 Cr 2 7 ? How many milligram-atoms of Cr 
does each milliliter of the dichromate contain? If 3.000 ml. of this dichromate 
o 2.000 ml. of KHC 2 O 4 .H 2 C 2 O 4 .2H 2 O solution =c= 1.000 ml. KOH o 3.000 ml. 
II 2 SO 4 , how many moles of Fe 2 3 .xII 2 O is each milliliter of the H 2 SO 4 capable 
of dissolving, and how many milliequivalents as an oxidizing agent would this 
amount of Fe 2 O 3 .xH 2 O represent? 

Ans. 4.467 grams. 0.2006 mg.-atoms. 0.00007514 moles, 0.1503 me. 

666. A sample of steel weighing 2.00 grams is analyzed for manganese by 
the bismuthate method. If a 25-inl. pipetful of 0.120 N FeSO 4 were used for 
the reduction of the oxidized manganese and 22.9 ml. of 0.0833 N KMnO 4 were 
used in the titration of the excess ferrous ions, what volume of the KMnO 4 
would have been used if the same weight of sample had been analyzed (a) by 
the chlorate method (using the same 25-inl. pipetful of the above FeSO 4 ); 
(6) by the Volhard method on a Y^ aliquot portion of the prepared solution? 
What is the percentage of Mn in the steel? 

Ans. (a) 30.8 ml, (6) 4.37 ml. 0.600 per cent. 

666. A sample of chrome iron ore weighing 0.3010 gram is fused with 
Na 2 O 2 , leached with water, and acidified with H 2 SO 4 . The resulting solution 
of dichromate is treated with a solution containing dissolved crystals of 



OXIDATION AND REDUCTION METHODS 231 

FeSO4.(NH 4 ) 2 SO4.6H 2 O, and the excess ferrous ions titrated with standard 
dichromate (containing 5.070 grams K 2 Cr 2 O7 per liter). A maximum of 
45.00 per cent Cr 2 O 3 in the ore being allowed for, what minimum weight of 
FeSO4.(NH 4 ) 2 SO4.6II 2 O should be used so that not more than a 50-ml. buretful 
of the standard dichromate would be required? 
Ans. 4.124 grams. 

667. A sample of pure sodium oxalate, Na2C 2 O 4 , weighing 0.2500 gram, when 
dissolved in dilute IT 2 SO 4 requires 40.15 ml. of eerie sulfate solution to give a 
permanent yellow color to the solution. What is the normality of the eerie 
sulfate solution? How many grams of pure Ce(SO4) 2 .2(NH4) 2 SO 4 .2H 2 O should 
be dissolved in 500 ml. of solution in order to prepare a solution of this nor- 
mality? If a sample of limonite weighing 0.3000 gram is dissolved in HC1, the 
iron reduced by metallic silver and then requires 25.03 ml. of the above eerie 
sulfate solution, orthophenaiithroline being used as indicator, what percentage 
of Fe 2 O 3 is shown to be present in the limonite? 

Ans. 0.09294 N. 29.40 grams. 62.03 per cent. 



668. What is the normality of a solution of KMnCX and what is the value 
of each milliliter in terms of grams of Fe if when titrating a 0.1000-gram sample 
of impure KNO 2 (which is oxidized to nitrate) the buret reading is one-half 
the percentage of N 2 O 3 in the sample? How many gram-atoms of Mn does 
each liter of the KMnO 4 contain? 

669. What must be the value of 1 ml. of eerie sulfate in terms of grams of 
Fe 2 O 3 so that in the titration of a half-gram sample of impure sodium arsenite 
(arsenite oxidized to arsenate), the percentage of As 2 O 3 in the sample will be 
twice the buret reading? What is the molarity of the eerie solution? 

660. A stock solution of KMnO 4 is made up and standardized. It is found 
that each milliliter is equivalent to 0.01597 gram of Fe 2 O 3 . A 10-ml. pipetful 
of the permanganate is reduced with H 2 O 2 in the presence of acid and the excess 
H 2 O 2 is destroyed by boiling. The resulting solution is then made neutral and 
the manganous ions in the solution are titrated with more of the original 
stock KMnO4, the solution being kept neutral with ZnO (Volhard method). 
How many milliliters of KMnC>4 would be required in the titration? 

661. A student standardized a solution of KOH and one of KMnO 4 against 
the same salt (KHC 2 O 4 .H 2 C 2 O4.2H 2 0). The normality of the former was 
found to be 0.09963 as a base and of the latter to be 0.1328 as an oxidizing 
agent. By coincidence, exactly 50.00 ml. of solution were used in each 
standardization. Calculate the ratio of the weight of tetroxalate used in the 
first case to that used in the second case. 

662. A powder is composed of oxalic acid (H 2 C 2 O 4 .2H 2 O), potassium 
binoxalate (KHC 2 O4.H 2 O), and an inert impurity. Find the percentage of 
each constituent from the following. A sample of the powder weighing 
1.200 grams reacts with 37.80 ml. of 0.2500 N NaOH solution; 0.4000 gram of 
powder reacts with 43.10 ml of 0.1250 N permanganate solution. 



232 CALCULATIONS OF ANALYTICAL CHEMISTRY 

663. It requires 15.27 ml. of SnCl 2 solution to reduce an amount of iron 
that can be oxidized by 16.27 ml. of permanganate solution. This volume 
of the permanganate will also oxidize that amount of KHC 2 O4.H 2 C 2 O4.2H 2 O 
solution which reacts with 16.24 ml. of 0.1072 N NaOH. Calculate the 
normality of the SnCl 2 solution. 

664. Given the following data, calculate the percentage of MnO 2 in a 
sample of pyrolusite: 

Sample = 0.5217 gram 

KHC 2 O4.H 2 C2O4.2II 2 O added to react with MnO 2 = 0.7242 gram 

KMiiCX used in titrating excess = 22.42 ml. 

1.000 ml. KMnO 4 ~ 0.009721 gram H 2 C 2 O 4 .2H 2 O 

666. A 50.00-ml. pipetful of 0.2016 N oxalic acid is added to a sample of 
pure MiiO 2 to reduce it. The excess of oxalic acid requires 10.15 ml. of 
0.2008 N KMnO 4 for its oxidation. What weight of MnO 2 is present? 

666. How many grams of Cr 2 3 are present in a sample of chromite ore if 
when decomposed by fusion with Xa 2 O 2 , acidified with H 2 SO 4 , and treated 
with 3.000 millimoles of KHC 2 O 4 .H 2 C 2 O 4 .2H 2 O(Cr 2 O 7 - + 3C 2 (V + 14H+ - 
2CV"*- 4 - + 6CQ* + 7H 2 O), the excess oxalate requires 20.00 ml. of 0.1000 X 
KMn0 4 ? 

667. A sample of spathic iron ore is analyzed for calcium by the perman- 
ganate method, following the precipitation of the calcium as oxalate. What 
weight of sample must be taken so that one-half the number of milliliters of 
0.1000 N KMnO 4 may represent the percentage of CaO in the sample? 

668. What weight of iron ore should be taken for analysis so that the 
milliliters of 0.0833 N permanganate multiplied by 2 will give the percentage 
of Fe 2 Oa in the sample? 

669. The qualitative analysis of a certain silicate shows the presence of a 
large quantity of calcium and only traces of other positive elements. In the 
quantitative analysis, the silica is removed and the calcium is precipitated 
from the filtrate as calcium oxalate. It is found that the milliliters of 0.1660 
X KMnOi required to oxidize the oxalate in a half-gram sample is almost 
exactly equal to the percentage of silica in the sample. What is the empirical 
formula of the pure mineral? 

670. Hculnnditc is hydrous acid calcium metasilicate and yields on analysis 
14.8 per cent water and 16.7 per cent alumina. If the calcium were precipitated 
as calcium oxalate from a 1.00-gram sample, 32.8 ml. of 0.100 X KMn0 4 
would be required for oxidation. Three-fifths of the water exists as water of 
crystallization. What is the empirical formula of heulandite? 

671. A sample of alloy containing manganese and weighing 4.35 grams is 
dissolved and the manganese eventually titrated in neutral solution with a 
standard permanganate having an "iron value" of 0.00640 gram (i.e., 1.000 ml. 
will oxidize that amount of ferrous iron in add solution). A volume of 13.05 
ml. is required. Calculate the percentage of Mn in the alloy. 

672. A sample of magnetite (impure FeaCW is fused with Xa 2 O 2 and all the 
iron thus oxidized to the ferric state. After leaching with water, the iron in 



UA1VAT1UJM AND KUDUUT1ON METHODS 233 

the solution is determined by reducing with SnCU, destroying the excess stan- 
nous ions, and titrating with 0.3000 N K 2 Cr 2 O 7 . If 30.00 ml. are required, 
calculate the number of grams of Fe 3 O 4 in the sample. How many grams of Cr 
are present in each milliliter of the K 2 Cr 2 07? How many milligrams of CeO 2 
is each milliliter of the dichromate equivalent to as an oxidizing agent? 

673. From the following data, compute the weight of iron ore to be taken 
for analysis such that the percentage of Fe 2 O 3 prasent is numerically equal to 
twice the number of milliliters of K 2 Cr 2 O 7 used in the titration. 

40.00 ml. HC1 solution o 2.880 grams of AgCl. 

35.00 ml. HC1 solution =c= 40.00 ml. of KHC 2 O 4 .H 2 C 2 O 4 solution. 

35.00 ml. of tetroxalate solution =0= 40.00 ml. of K 2 Cr 2 O 7 solution. 

674. A solution of dichromate is prepared by dissolving 4.883 grams of 
pure K 2 Cr 2 O 7 and diluting to exactly one liter; a solution of ferrous salt is pre- 
pared by dissolving 39.46 grams of FeSO 4 .(NH4) 2 SO 4 .6H 2 O and diluting to 
one liter. What volume of the dichromate solution must be transferred to the 
ferrous solution and thoroughly mixed so that the normality of one solution as 
a reducing agent will be the same as the normality of the other solution as an 
oxidizing agent? 

676. An oxide of iron weighing 0.1000 gram is fused with KHSO 4 , and the 
fused material is dissolved in acid. The iron is reduced with stannous chloride, 
mercuric chloride is added to oxidize the excess stannous ions, and the iron is 
titrated with 0.1000 N dichromate solution. If 12.94 ml. were used, what is the 
formula of the oxide FeO, Fe 2 O 3 , or Fe 3 O 4 ? 

676. Two millimoles of pure Pb 3 O 4 (= PbO 2 .2PbO) are dissolved in 
a solution containing a mixture of 25 ml. of 6 N H 2 SO 4 and A grams of 
FeSO 4 .(NH 4 ) 2 SO 4 .6H 2 O, the 4-valent lead being reduced to Pb++. The 
excess of ferrous ions requires 12.00 ml. of 0.2500 N KMnO 4 for oxidation 
(a) What is the value of Al (6) How many milligram-atoms of Mn are 
present in each milliliter of the KMnO 4 ? (c) If potassium tetroxalate, 
KHC 2 O 4 .H 2 C 2 O 4 .2H 2 O, had been substituted for the ferrous ammonium sulfate 
above, how many milliequivalents, how many millimoles, and how many 
grams of the oxalate would have been used for the reduction so that the excess 
oxalate would have required 12.00 ml. of 0.2500 N KMnO 4 ? (d) If lead 
sesquioxide, Pb 2 O 3 , were analyzed by a similar method, what would be the 
milliequivalent weight of Pb 2 O 3 ? 

677. From the following data, calculate the percentage of iron in a sample 
of limonite: 

1.000 ml. K 2 O 2 O 7 o 0.006299 gram Fe 
Dichromate solution used = 47.56 ml. 

Ferrous solution used = 2.85 ml. 
Sample taken for analysis = 0.6170 gram 
1.000 ml. ferrous solution =0= 1.021 ml. K 2 O 2 O 7 

678. A certain chrome iron ore is known to contain 24.80 per cent Cr. 
A sample weighing 0.2580 gram is fused with Na 2 O 2 , leached with water, and 
acidified with H 2 SO 4 . The resulting solution of dichromate is treated with a 



234 CALCULATIONS OF ANALYTICAL CHEMISTRY 

weight of FeSO4.7H 2 O crystals which happens to be just 50 per cent more than 
the amount necessary to reduce the dichromate. The excess of ferrous ions is 
titrated with standard dichromate (containing 0.02000 millimole K 2 Cr 2 O7 per 
milliliter). What volume is required? (Hint: It is not necessary to calculate 
the amount of ferrous salt required.) 

679. A solution of eerie sulfate is of such normality that 26.73 ml. 
are required to titrate the ferrous iron obtainable from 1.052 grams of 
FeSO 4 .(NH4)2SO4.6H 2 O. How many grams of Ce(S0 4 ) 2 .2(NH 4 ) 2 SO 4 .2H 2 O 
should be dissolved in 750 ml. of water and the resulting solution diluted to 
one liter in order to prepare a solution of such normality that each milliliter 
is equivalent to 0.006500 gram KHC 2 O 4 .Il2C 2 O 4 .2H 2 O? 

680. (a) What is the normality of a solution of KMn0 4 if each milliliter 
will oxidize 0.008377 gram of iron from the ferrous to the ferric state? (b) How 
many grams of Mn do 10.00 ml. of such a solution contain? (c) How many 
grams of Mn would 10.00 ml. of this KMnO 4 oxidize to MnO 2 by the Volhard 
method (3Mn++ + 2MnO 4 ~ + 2ZnO -> 5MriO 2 + 2Zn- f +)? (d) How many 
milliliters of this KMn0 4 would be required to titrate 0.1200 millimole of 
sodium formate (NaCIIO 2 ) according to the reaction: 3CHO 2 ~ + 2MnO 4 ~ + 
H 2 O -> 2MnO 2 + 3CO 2 + 50H-? (e) How many grams of CaO would each 
milliliter of the KMnO 4 be equivalent to in the volumetric method for calcium? 
(/) How many grams of As 2 O 3 would each milliliter of the KMnO 4 be equiva- 
lent to in the titration of arsenite to arsenate? 

681. Find the percentage of Pb 3 4 in a sample of red lead that has been 
adulterated with PbO. 2.500 grams of the pigment are treated with 50.00 ml. 
of potassium tetroxalate solution which is 0.1500 N as an acid, and the excess 
of the latter requires 30.00 ml. of permanganate of which each milliliter is 
equivalent to 0.005584 gram of iron. 

682. A sample of steel weighing 2.50 grams is analyzed for manganese by 
the chlorate method (see above). If a 25-ml. pipetful of 0.110 N FeSO 4 were 
used to dissolve the precipitated Mn0 2 and 18.4 ml. of 0.125 N KMnO 4 were 
used to titrate the excess ferrous ions, what volume of the permanganate 
would have been used if the same weight of sample had been analyzed (d) by 
the bismuthate method (using the same 25-ml. pipetful of the above FeSO 4 ) ; 
(b) by the Volhard method on a J^ aliquot portion of the prepared solution? 
What is the percentage of Mn in the steel? 

94. lodimetric Process. The fundamental reaction in this 
process is that between iodine and sodium thiosulfate, with starch 
(or sometimes chloroform) as the indicator. 

I 2 + 28203= -> 21- + S^- 

Titrations by this process may be divided into two groups, those 
involving direct titrations with standard iodine and those involv- 
ing titrations with standard sodium thiosulfate. 



OXIDATION AND REDUCTION METHODS 235 

Iodine solutions are prepared by dissolving iodine crystals, to- 
gether with potassium iodide, in water. A normal solution con- 
tains I 2 /2 = 126.9 grams of iodine per liter. Standard solutions 
are used to titrate directly certain reducing agents of which the 
following are typical: 

SUBSTANCE OXIDIZED TO 

H 2 S S 

so 3 ~ sor 

S 2 o 3 - s 4 o c - 



Sodium thiosulfate solutions are prepared by dissolving crystals 
of the salt in water. A normal solution contains Na2S2O3.5Il20/l = 
248.2 grams of the hydrated salt per liter (see Sec. 89). Stand- 
ard solutions of thiosulfate can be used to titrate almost any 
oxidizing substance. The titration is made, however, by adding 
to the solution of oxidizing substance a large excess (roughly 
measured) of potassium iodide. The oxidizing substance is re- 
duced, liberating an equivalent amount of iodine, and the liberated 
iodine is titrated with thiosulfate. Typical oxidizing agents de- 
termined in this way are as follows: 

SUBSTANCE EQUATION 

Cr 2 O 7 ~ O 2 O 7 - + 61- + 14H+ -> 2Cr + + + + 3I 2 + 7H 2 O 

Mn0 4 - 2MnO 4 - + 10I~ + 16H+ -> 2Mn++ + 5I 2 + 8H 2 O 

BKV BrO 3 - + 61- + 6H+ -> Br~ + 3I 2 + 3H 2 O 

IOr I0r + 61- + 611+ -> I- + 3I 2 + 3II 2 O 

Cu++ 2Cu++ + 41- -> Cu 2 I 2 + I 2 



C1 2 C1 2 + 21- -> 2C1- + I 2 

II 2 O 2 H 2 O 2 + 21- + 211+ - I 2 + 2H 2 O 

Since an oxidizing agent liberates its own equivalent of iodine, 
the volume of thiosulfate required for the liberated iodine in any 
given case is the same as would be required if the thiosulfate were 
used directly and reduced the substance to the form indicated. 
In calculations, therefore, the equivalent weight of the substance 
titrated is found in the usual way by dividing the formula weight 
of the substance by the total change in oxidation number. 

In titrations in acid solution, a standard solution of potassium 
iodate containing an excess of potassium iodide is a convenient 
substitute for standard iodine. It is a colorless, stable solution, but 



236 CALCULATIONS OF ANALYTICAL CHEMISTRY 

when it comes in contact with acid the two ingredients immedi- 
ately interact and liberate free iodine (IO 3 ~~ + 61"" j- 6H+ I~ + 
3I 2 + 3H 2 0). In the titration of a substance in acid solution, this 
standard solution, therefore, behaves as if it were a standard solu- 
tion of iodine. It is used, for example, in the determination of 
sulfur in steel. Since the iodate molecule has the oxidizing 
equivalent of 6 iodine atoms, a tenth-normal solution contains 
KI0 3 /60 = 3.567 grams of KIO 3 per liter and can be prepared 
by dissolving this amount of the pure crystals, together with 
an excess of potassium iodide, in water and diluting to exactly one 
liter. 

EXAMPLE I. An excess of potassium iodide is added to a solu- 
tion of potassium dichromate, and the liberated iodine is titrated 
with 48.80 ml. of 0.1000 N sodium thiosulfate solution. How 
many grams of K 2 Cr 2 07 did the dichromate solution contain? 

SOLUTION: Potassium dichromate liberates an equivalent 
amount of iodine from an iodide (i.e., 6 gram-atoms = 6 gram- 
equivalents of iodine per mole of dichromate) : 

Cr 2 0r + 61- + 14H+ -> 2Cr+++ + 3I 2 + 7H 2 O 
and the liberated iodine is titrated with thiosulfate 

2S 2 O 3 ~ + k -> S 4 O 6 ~ + 21- 

The volume of titrating solution is the same as it would have been 
if the original solution had been titrated directly to the indicated 
products. 



Grams of K 2 Cr 2 O 7 = 48.80 X 0.1000 X 7 - 0.2393. An*. 



EXAMPLE II. The sulfur from 4.00 grams of steel is evolved 
as H 2 S and titrated with 1.60 ml. of 0.05000 N iodine solution. 
What is the percentage of S in the steel? What is the value of 
1.000 ml. of the iodine in terms of As 2 O 3 ? How many milliliters 
of the iodine will be reduced by 40.00 ml. of Na 2 S 2 3 solution of 
which 1.000 ml. =c= 0.006357 gram Cu. What volume of iodate- 
iodide solution containing 10.0 millimoles of KIO 3 and 50.0 grams 
of KI per liter would be required to titrate the H 2 S from 5.00 grams 
of the above steel? The equations involved are as follows: 



OXIDATION AND REDUCTION METHODS 237 



1 + 2HCOr + !- As0 4 s + 21- + 2CO 2 + H 2 O 

2Cu++ + 41- -> Cu 2 I 2 + I 2 
IO 3 - + 61- + 6H+ -* 3I 2 + I- 4- 3H 2 O 
SOLUTION: 

1. 60 X 0.05000 Xn-TjL 

' X 100 = 0.0320 per cent S. Ans. 

1.000 X 0.05000 X -r^ = 0.002473 gram AszO 3 . Ans. 



The addition of KI to a copper solution will cause reduction of 
the Cu and the liberation of an amount of iodine equivalent to 
the copper present. This iodine may be titrated with thiosulfate 
and the normality of the latter found from the amount of Cu 
present. In the above case, 

Normality of Na 2 S 2 O 3 solution = | 000 X Cii/1 000 = ' 1000 N 

1000 
Volume of 0.0500 N I 2 solution = 40.00 X = 80.00 ml. 



Ans. 

10.0 millimoles KIOs = 60.0 milliequivalents 
Normality of KIO, = 0.0600 N 



x = 1.67 ml. Ans. 

The reaction As0 8 = + I 2 + H 2 O <= AsOr + 21- + 211+ is revers- 
ible, for 3-valent arsenic is oxidized by iodine in neutral solution, 
whereas 5-valent arsenic is reduced by iodide in the presence of 
acid with the liberation of free iodine. These reactions can be 
made use of in the determination of the two forms of arsenic when 
present in the same solution. 

EXAMPLE III. A powder consists of Na 2 HAs0 3 + As 2 O 5 + inert 
material. A sample weighing 0.2500 gram is dissolved and titrated 
with standard iodine in a solution kept neutral by excess dissolved 
NaHCO 3 (AsO 3 s + I 2 + 2HCO 3 - - AsO 4 s + 21- + 2CO 2 + H 2 O). 
The titration requires 15.80 ml. of 0.1030 N I 2 . Hydrochloric acid 
and an excess of KI are added (AsO<r + 21- + 8H+ - AS+++ -f 



238 CALCULATIONS OF ANALYTICAL CHEMISTRY 

I 2 + 4H 2 O) and the liberated iodine requires 20.70 ml. of 0.1300 
N NajSaOs. Calculate the percentages of Na 2 HAsOs and 
in the sample. 
SOLUTION: 

15.80 X 0.1030 = 1.628 milliequivalents 3-valent As 
20.70 X 0.1300 = 2.691 milliequivalents total As 
2.691 - 1.628 = 1.063 milliequivalents 5-valent As 

1.628 X 



9 ono 

X 100 = 55.33 per cent 



0.2500 , A 

> Ans. 

! 063 X As2 5 

' X 100 = 24.42 per cent As 2 B 

U.^oUU 

Problems 

683. A solution of iodine contains 15.76 grams of I 2 per liter. What is 
the value of each milliliter as an r oxidizing agent in terms of (a) SO 2 , (6) H 2 SOj, 
(c) Na 2 S 2 O 3 , (d) As? 

Ans. (a) 0.003978 gram, (6) 0.005097 gram, (r) 0.01963 gram, (d) 0.004655 
gram. 

684. What is the value of 1.000 ml. of 0.04000 N sodium thiosulfate solution 
in terms of Cu? What is the normality of a thiosulfate solution if 25.00 ml. 
are required to titrate the iodine liberated by 0.01563 gram of copper? 

Ans. 0.002543 gram. 0.009833 N. 

686. What is the value of 1.000 ml. of an iodine solution (1.000 ml. o= 
0.03000 gram Na 2 S 2 O 3 ) in terms of As 2 O 8 ? 
Ans. 0.009386 gram. 

686. From the following data calculate the normality and molarity of the 
Na 2 S 2 O 3 solution and the value of 1.000 ml. in terms of grams of potassium 
bi-iodate[KH(I0 8 )l. 

1.000 ml. K 2 Cr 2 7 o 0.005585 gram of Fe. 

20.00 ml. K 2 Cr 2 O 7 liberates sufficient iodine from potassium iodide to require 
32.46 ml. Na 2 S 2 3 solution for reduction. 

Ans. 0.06162 N, 0.06162 M, 0.002003 gram. 

687. Forty milliliters of KMnO 4 solution (1.000 ml. <^ 0.005000 gram Fe) 
are added to KI and the liberated iodine is titrated with sodium thiosulfate 
solution requiring 35.90 ml. What is the value of 1.000 ml. of the thiosulfate 
solution in terms of copper? 

Ans, 0.006345 gram. 

688. A solution of sodium thiosulfate is freshly prepared, and 48.00 ml. are 
required to titrate the iodine liberated from an excess of KI solution by 



OXIDATION AND REDUCTION METHODS 239 

0.3000 gram of pure KIOs. What are the normality of the thiosulfate and its 
value in terms of iodine? 

Ans. 0.1752 N, 0.02224 gram. 

689. The thiosulfate solution of the preceding problem is allowed to stand, 
and 1.00 per cent of the Na 2 S 2 O 3 is decomposed by a trace of acid present in 
the solution (S 2 O 3 = + 2H+ -> H 2 SO 3 + S). What is the new normality of 
the solution as a reducing agent, assuming oxidation of sulfite to sulfate? 

Ans. 0.1768 N. 

690. A steel weighing 5.00 grams is treated with NCI and the H 2 S is evolved 
and eventually titrated with a solution containing 0.0100 mole of KIO 3 and 
60 grams of KI per liter. If 3.00 ml. are required, what is the percentage of 
sulfur in the steel? 

Ans. 0.0576 per cent. 

691. If 20.00 ml. of thiosulfate (1.000 ml. ^ 0.03750 gram CuSO 4 .5H 2 O) 
are required for a certain weight of pyrolusite by the Bunsen iodimetric method, 
what weight of H 2 C 2 O 4 .2H 2 O should be added to a similar sample to require 
20.00 ml. of 0.1000 N KMnO 4 by the commonly used indirect method? In 
the Bunsen method MnO 2 is reduced by TTC1, the C1 2 liberated from the 
latter is passed into KI solution, and the liberated iodine is titrated with 
thiosulfate. 

Ans. 0.3154 gram. 

692. Titrating with 0.05000 N iodine, what weight of stibnite ore should be 
taken so that the percentage of Sb 2 S 3 in the sample will be 1 J^ times the buret 
reading? (SbOr + I 2 + 2HCO 3 - -+ SbCV + 21- + 2CO 2 + H 2 O). How 
many millimoles of Na 2 S 2 O 3 .5H 2 O is each liter of the above iodine equivalent to? 

Ans. 0.2831 gram. 50.00 millimoles. 

693. A sample of sodium sulfite weighing 1.468 grams was added to 
100 ml. of 0.1000 N iodine. The excess iodine was titrated with 42.40 ml. of 
Na 2 S 2 O 3 .5H 2 O solution of which 1.000 ml. was equivalent to the iodine liberated 
from 0.01574 gram of KI. Calculate the percentage of Na 2 SO 3 in the sample. 

Ans. 25.67 per cent. 

694. A sample of stibnite containing 70.00 per cent Sb is given to a student 
for analysis. He titrates with a solution of iodine of which he had found 
1.000 ml. to be equivalent to 0.004948 gram of As 2 O 3 . The normality of the 
solution, however, had changed, owing to volatilization of iodine, and the 
student reports 70.25 per cent Sb. What are the percentage error and the 
present normality of the iodine solution, and how much 0.2000 N iodine solu- 
tion must be added to one liter of the solution to bring it back to its original 
strength? (SbOr + I 2 + 2HCOr -> SbO 4 3 + 21- + 2CO 2 + H 2 O.) 

Ans. 0.36 per cent, 0.09965 N. 3.5 ml. 

696. A sample of impure potassium iodide weighing 0.3100 gram is treated 
with 0.1942 gram ( = 1 millimole) of K 2 CrO 4 and 20 ml. of 6 N H 2 SO 4 and the 
solution is boiled to expel all the free iodine formed by the reaction. The 



240 CALCULATIONS OF ANALYTICAL CHEMISTRY 

solution containing the excess chromate is cooled, treated with excess KI, and 
the liberated I 2 titrated with 0.1000 N Na 2 S 2 O 3 , requiring 10.20 ml. Calculate 
the percentage purity of the original potassium iodide. 
Ans. 96.38 per cent. 

696. A standard stock solution is made by dissolving 50.0 millimoles of 
KIO 3 and 100 grams of KI in water and diluting to 10.00 liters. A sample 
of Bureau of Standards steel weighing 5.00 grams and certified to contain 
0.0530 per cent sulfur is treated with HC1. The sulfur is liberated as H 2 S, 
caught in ammoniacal zinc sulfate solution, and eventually titrated in the 
presence of acid with the above-mentioned stock solution. What volume is 
required? What is the oxidizing normality of the iodate solution? How 
many grams of KI in excess of the theoretical amount were used in preparing 
the stock solution? 

Ans. 5.52 ml. 0.0300 N. 58.5 grams. 

697. The copper in a 0.2500-gram sample of copper ore is treated in 
solution with excess KI and the liberated iodine titrated with 16.50 ml. of 
Na^Oa (1.000 ml. ^ 0.003619 gram KBrO 3 ). What is the purity of the ore 
expressed in terms of percentage of Cu 2 S? 

Ans. 68.28 per cent. 

698. If the arsenic trichloride from 50 grams of impure copper is distilled 
off, absorbed in dilute alkali, and finally oxidized by 20 ml. of 0.0200 N 
iodine solution, find the percentage of arsenic in the sample. 

Ans. 0.030 per cent. 

699. A mixture of As 2 O 3 and As 2 O 6 and inert matter is dissolved and titrated 
in neutral solution with 0.05000 N I 2 , requiring 20.10 ml. The resulting solu- 
tion is acidified and excess KI is added. The liberated I 2 requires 29.92 ml. of 
0.1500 N Na 2 S 2 3 . Calculate the number of grams of combined As 2 O 3 + As 2 O 6 
in the sample. 

Ans. 0.2498 gram. 

700. A mixture of pure potassium permanganate and pure potassium 
chromate weighing 0.2400 gram, when treated with KI in acid solution, 
liberates sufficient iodine to react with 60.00 ml. of 0.1000 N sodium thio- 
sulfate solution. Find the percentages of Cr and Mn in the mixture. 

Ans. Cr = 11.0 per cent, Mn = 20.5 per cent. 

701. A sample of pyrolusite is treated with ITC1, the liberated chlorine is 
passed into potassium iodide, and the liberated iodine is titrated with Na 2 S 2 O 3 
solution (49.64 grams of Na 2 S 2 O 3 .5H 2 O per liter). If 38.70 ml. are required, 
what volume of 0.2500 N KMnO 4 would be required in an indirect determina- 
tion in which a similar sample is reduced with 0.9000 gram of H^CaO^I^O 
and the excess oxalic acid is titrated with the KMnO 4 ? 

Ans. 26.16ml. 

702. One milliliter of a thiosulfate solution is equivalent to 0.005642 gram 
of copper and is also equivalent to 1.50 ml. of a certain iodine solution. Cal- 



OXIDATION AND REDUCTION METHODS 241 

culate the value of one milliliter of the iodine solution in terms of grams of 
(a) Sb, (b) As, (c) As 2 S 8 , (rf) Sb 2 O 8 . 

703. Calculate the percentage of iron in a sample of crude ferric chloride 
weighing 1.000 gram if the iodine liberated by its action on an excess of potas- 
sium iodide is reduced by the addition of 50.00 ml. of Na 2 S 2 O 3 solution and the 
excess thiosulphate is titrated with standard iodine, of which 7.85 ml. are 
required. 

45.00 ml. iodine =0= 45.95 ml. thiosulfate 
45.00 ml. arsenite solution =c= 45.20 ml. iodine 
1.000 ml. arsenite solution o 0.00516 gram As 2 C>3. 

704. What weight of copper ore should be taken for analysis so that when 
the copper is determined by the regular iodimetric method using 0.05000 N 
sodium thiosulfate, the buret reading will be two-thirds the per cent CuS in 
the ore? What is the molarity of the thiosulfate solution? What volume of 
KMnO 4 (1.000 ml. =0= 0.0005493 gram of Mn by the Volhard method: 
3Mn++ + 2MnO 4 - + 2ZnO -> 5Mn0 2 + 2Zn++ + 2II 8 O) will react with an 
excess of soluble iodide in the presence of acid to require 20.00 ml. of the above 
thiosulfate for reduction? 

706. A special analysis calls for the preparation of an exactly 0.1000 N 
solution of sodium thiosulfate. The usual method is to prepare a solution 
of slightly greater strength and standardize it, subsequently diluting with a 
calculated and carefully measured amount of water. In this case a stand- 
ardization of the diluted solution was also made to ensure its accuracy. 
The data follow: 

FIRST STANDARDIZATION SECOND STANDARDIZATION 

Pure Cu, grams Thiosulfate, ml. Pure Cu, grams Thiosulfate, ml. 

0.2504 37.98 0.2492 39.09 

0.2592 39.24 0.2507 39.27 

0.2576 39.04 0.2631 41.28 

Calculate the volume of water necessary to reduce 1,000 ml. of the first 
solution to a normality of 0.1000, and calculate the amount that apparently 
was actually added. 

706. If 50.00 ml. of an iodine solution are exactly equivalent in oxidizing 
power to 49.47 ml. of a K 2 Cr 2 O 7 solution of which 1.000 ml. will liberate 
0.004263 gram of iodine from KI, calculate the normality of each solution. 

707. Pure K 2 Cr 2 C>7, weighing 0.3321 gram, was boiled with an excess of 
strong HC1. The evolved chlorine was passed into a soluton of KI, and the 
I 2 liberated was titrated with 68.21 ml. of Na2S2O 3 solution. Calculate the 
normality of the Na 2 S 2 O 3 solution. 

708. What must be the normality of a standard iodine solution so that, 
if a 0.5000-gram sample of stibnite (impure Sb 2 Ss) is taken for analysis, the 
number of milliliters of solution may represent directly the percentage of 
antimony? (SbO 8 - + It + H 2 O -> SbO 4 " + 2H+ + 2I~). 



242 CALCULATIONS OF ANALYTICAL CHEMISTRY 

709. A powder consists of a mixture of Na 3 AsO4.12H 2 O, Na2HAsO 3 , and 
inert matter. It is dissolved and titrated in neutral solution with 0.08100 N I 2 , 
requiring 15.60 ml. The resulting solution is acidified and excess KI is added. 
The liberated iodine requires 8.58 nil. of 0.1200 molar Na 2 S 2 O 3 solution. 
Calculate the amount of 5-valent and of 3-valent arsenic in terms of grams of 
Na 3 AsO 4 .12H 2 O and grams of combined As 2 O 3 , respectively. 

710. What weight of sulfite liquor should be taken for analysis so that 
the milliliters of I 2 solution required (1.000 ml. - 0.0125 gram Na 2 S 2 O 3 .5II 2 O) 
and the percentage total SO 2 shall be in the respective ratio of 20:3? 

711. If the amount of copper in a carbonate ore, expressed in terms of 
percentage Cu 2 (OIT) 2 CO 3 is 53.05 and if 25.72 ml. of Na 2 S 2 O 3 are eventually 
required to titrate the iodine liberated from excess KI by the copper from a 
half-gram s'ample, what is the value of 1.000 ml. of the thiosulfate in terms of 
grams of (a) KBrO 3 , (b) KII(IO 3 ) 2 ? 

712. The sulfur from a 5.00-gram sample of steel is evolved as II 2 S and 
eventually titrated in the presence of acid with standard iodine solution 
(1.00 ml. o 0.004945 gram As 2 O 3 ), of which 1.90 ml. are required. What is the 
percentage of sulfur in the steel? If a standard potassium iodate-iodide 
solution had been substituted for the standard iodine and a volume identical 
with the above had been required, how many grams of KIO 3 would have 
been present in each milliliter of the solution? Could the iodate solution 
have been standardized against pure As 2 O 3 , as the iodine solution was? 
Explain your answer. 



CHAPTER XIV 

PRECIPITATION METHODS 
(PRECIPITIMETRY) 

95. Equivalent Weights in Precipitation Methods. In precipi- 
tation (or "saturation") methods a substance is titrated with a 
standard solution of a precipitating agent. At the completion of 
the precipitation the precipitating agent reacts with an indicator 
and a color change takes place. For example, in the Volhard 
method for silver, silver ions are titrated with a standard solution 
of potassium thiocyanate (Ag+ + CNS~ AgCNS) and the end 
point is determined by the red color formed when an additional 
drop of the thiocyanate reacts with ferric alum indicator (Fe* 4 " 1 " + 
4CNS--*Fe(CNS) 4 -). 

Similarly silver ions and halide ions can be titrated in neutral 
solution with standard NaCl or standard AgNO 3 using certain so- 
called adsorption indicators (e.g., fluorescein), which form colored 
compounds on the surface of the particles of precipitate and give 
a change of color at the equivalence point. 

In determining the equivalent weight of a constituent being 
precipitated, 1.008 grams of hydrogen ion is again taken as the 
stadnard of reference. The equivalent weight is that weight which 
in precipitation reacts with the equivalent of that amount of hy- 
drogen ion. Here the point of view is that of general metathesis 
rather than that of neutralization (as in acidimetry) or of oxidizing 
power (as in oxidimetry). Knowledge of valence and a simple 
inspection of the equation usually suffice to determine the correct 
equivalent weight. In a great majority of cases the gram-equiva- 
lent weight is found by dividing the formula weight by the net 
number of charges on the constituent actually taking part in the 
reaction. 

In the reaction between silver nitrate and sodium chloride 
(Ag+ + Cl~ AgCl) the equivalent weights of the reacting sub- 
stances are AgNOs/1 and NaCl/1, respectively. In the reaction 
between barium chloride and sodium sulfate (Ba ++ + SO 4 ==: > 

243 



244 CALCULATIONS OF ANALYTICAL CHEMISTRY 

BaSO4) the equivalent weight in each case is one-half of the mo- 
lecular weight. The equivalent weight of anhydrous disodium 
phosphate as an acid is Na2HPC>4/l = 142.05; as a sodium salt, 
Na2HPO 4 /2 = 71.03; and as a phosphate, Na 2 HP0 4 /3 = 47.35. 

Reactions in this class may be direct or indirect. That is, the 
titrating solution may be added in amounts just sufficient to pre- 
cipitate all of the substance to be determined; or in certain cases 
an excess of the precipitating agent may be added and the excess 
titrated by means of a precipitating agent. Because of the diffi- 
culty of finding suitable indicators to show the completion of the 
reactions a great many precipitating reactions cannot be used 
satisfactorily for quantitative titrations. 

The Volhard method for silver is a direct method and is illus- 
trated in Example I below. The Volhard method can also be 
applied as an indirect process to the determination of chloride, 
bromide, iodide, cyanide, and thiocyanate. This is illustrated in 
Example II below. 

EXAMPLE I. What is the percentage of silver in a coin, if a 
0.2000-gram sample requires 39.60 ml. of potassium thiocyanate 
solution (0.4103 gram of KCNS per 100 ml.) for the precipitation 

of the silver? 

Ag++CNS~~*AgCNS 

SOLUTION: A liter of the KCNS solution contains 4.103 grams 
of the salt. Its normality is 

4.103 4.103 



KCNS/1 97.17 
Ag 



0.04223 



39.60 X 0.04223 X 

nqnnft ^^ X 100 = 90.20 per cent Ag. Ans. 

u.^uuu 

EXAMPLE II. A sample of impure strontium chloride weighs 
0.5000 gram. After the addition of 50.00 ml. of 0.2100 N AgNO 3 
and filtering out of the precipitated silver chloride, the filtrate 
requires 25.50 ml. of 0.2800 N KCNS to titrate the silver. What 
is the percentage of SrCU in the sample? 
SOLUTION: 

Milliequivalents of AgNOs added - 50.00 X 0.2100 - 10.50 

Milliequivalents of KCNS required = 25.50 X 0.2800 = 7.14 

Net milliequivalents = 10.50 - 7.14 = 3.36 



PRECIPITATION METHODS 245 



3.< 
Per cent SrCl 2 - . Kt ^' w X 100 = 53.3 per cent. 

U.OUv/U 

EXAMPLE III. A sample of feldspar weighing L500 grams is 
decomposed, and eventually there is obtained a pure mixture of 
KC1 and NaCl weighing 0.1801 gram. These chlorides are dis- 
solved in water, a 50-ml. pipetful of 0.08333 N AgNO 3 is added, 
and the precipitate filtered off. The filtrate requires 16.47 ml. of 
0.1000 N KCNS, ferric alum being used as indicator. Calculate 
the percentage of K^O in the silicate. 

Let x = grams of KC1 obtained 

Then 

0.1801 - x - grams of NaCl 

Total milliequivalents of mixed halides = Trri/1 QAQ + 

0.1801-x 
NaCl/1,000 

(IG^ 
x - 0.1517 gram KC1 



so/boo + mm - (5 x - 08333> ~ (16 - 47 x - 1000) 



0.1517 



X K 2 O/2KC1 ^ inn A , A , A 

* ' - X 100 = 6.40 per cent. Ans. 
l.oOu 



Problems 

713. What volume of 0.1233 N silver nitrate solution is required to pre- 
cipitate the chlorine from a sample of rock salt weighing 0.2280 gram and 
containing 99.21 per cent NaCl and no other halide? 

Ans. 31.37 ml. / 

714. What volume of 0.08333 N BaCl 2 solution is required to precipitate 
the sulfur from a solution containing 0.2358 gram of FeSO 4 .7H 2 O? 

Am. 20.36 ml. 

715. A solution of a soluble phosphate that is 0.2000 N as a precipitating 
agent is used to precipitate the magnesium as MgNH 4 PO4 from a 1.000-gram 
sample of dolomite containing 14.01 per cent MgCO 3 . What volume is required? 

Ans. 16.62 ml. 

716. A solution of K 2 Cr 2 7 which is 0.1121 normal as an oxidizing agent 
is used to precipitate BaCrO 4 from 0.5060 gram of BaCl 2 .2H 2 O. What is the 
normality of the solution of K 2 Cr 2 C>7 as a precipitating agent, and what volume 
is required? 

Ans. 0.07473 N, 55.42 ml. 



246 CALCULATIONS OF ANALYTICAL CHEMISTRY 

717. What volume of oxalic acid solution which is 0.2000 N as an acid is 
required to precipitate the calcium as CaC 2 O4.H 2 O from 0.4080 gram of 
cement containing 60.32 per cent CaO? What is the normality of the oxalic 
acid as a precipitating agent? 

Ans. 43.88ml. 0.2000 N. 

718. In the volumetric analysis of a silver coin containing 90.00 per cent Ag, 
a 0.5000-grarn sample being used, what is the least normality that a potas- 
sium thiocyanate solution may have and not require more than 50.00 ml. of 
solution in the analysis? 

Ans. 0.08339 N. 

719. Pure elementary arsenic weighing 0.1500 gram is dissolved in HNO 3 
(forming HaAsO^. The resulting solution is made neutral and then treated 
with 150 ml. of 0.06667 M AgNO* which precipitates all the arsenic as Ag 3 AsO 4 . 
The precipitate is washed, dissolved in acid, and the silver in the resulting 
solution of the precipitate is titrated with 0.1000 M KCNS using ferric ions as 
indicator. How many milliliters are required? 

Ans. 60.06 ml. 

720. What is the percentage of bromine in a sample of bromide if to 1.600 
grams of the sample are added 52.00 ml. of 0.2000 N AgNO 8 solution and the 
excess silver requires 4.00 ml. of 0.1000 N KCNS solution for the precipitation 
of AgCNS? 

Ans. 49.95 per cent. 

721. The purity of soluble iodides is determined by precipitating the iodine 
with an excess of standard silver nitrate and titrating the excess AgNO 3 
with thiocyanate solution. The silver nitrate is made by dissolving 2.122 
grams of metallic silver in nitric acid, evaporating just to dry ness, dissolving 
the residue in water, and diluting to exactly 1,000 ml. From a buret 60.00 ml. 
of this solution are added to 100.0 ml. of a solution of an iodide and the excess 
is titrated with 1.03 ml. of thiocyanate solution of which 1.000 ml. will pre- 
cipitate 0.001247 gram of silver as AgCNS. Calculate the grams of iodine 
present as iodide in the 100-ml. portion of the solution. 

Ans. 0.1482 gram. 

722. A mixture of pure LiCl and Bal2 weighing 0.6000 gram is treated 
with 45.15 ml. of 0.2000 normal AgNO 3 solution, and the excess silver is then 
titrated with 25.00 ml. of 0.1000 N KCNS solution with ferric alum as an 
indicator. Find the percentage of iodine present in the mixture. 

Ans. 44.61 per cent. 

723. A sample of feldspar contains 7.58 per cent Na 2 O and 9.93 per cent 
K 2 0. What must be the normality of a silver nitrate solution if it takes 22.71 
ml. of it to precipitate the chloride ions from the combined alkali chlorides 
from a 0.1500-gram sample? 

Ans. 0.03005 N. 



PRECIPITATION METHODS 247 

724. A sample of greensand weighing 2.000 grams yields a mixture of NaCl 
and KC1 weighing exactly 0.2558 gram. After the chlorides have been dis- 
solved, 35.00 ml. of 0.1000 N AgNO 3 are added to precipitate the chlorine 
and the excess is titrated with 0.92 ml. of 0.02000 N thiocyanate solution. 
Calculate the percentage of potassium in the sample. 

Ans. 6.36 per cent. 

726. A 1.000-gram sample of feldspar containing 3.05 per cent Na^O and 
10.0 per cent K 2 O is decomposed with CaCO 3 + NH 4 C1 and eventually yields 
a residue of NaCl + KC1 which is treated with 75.0 ml. of 0.06667 M AgNO 8 . 
The precipitate is filtered off and the filtrate titrated with 0.1000 M KCNS. 
How many milliliters are required? 

Ans. 18.93 ml. 



726. How many milliliters of 0.2500 N AgNO 3 solution are required to 
precipitate all the chlorine from a solution containing 0.5680 gram of 
BaCl 2 .2H 2 0? 

727. How many milliliters of a solution of Na 2 HPO4.12H 2 O which is tenth 
normal as a sodium salt are required to precipitate the calcium as Ca 3 (PO 4 )2 
from a solution containing 0.5000 gram of Ca(NO 3 ) 2 ? 

728. How many milliliters of K 2 Cr 2 O 7 (1.000 ml. 0.01597 gram Fe 2 O 3 ) 
will precipitate all the lead as PbCrO 4 from a solution containing 0.2510 gram 
of Pb(N0 3 ) 2 ? 

729. To precipitate the sulfur from a certain weight of ferrous ammonium 
sulfate contaminated with silica and water requires a number of milliliters 
of 0.2000 N barium chloride solution exactly equal to the percentage of iron 
in the sample. What is the weight of sample? 

730. In the analysis of a sample of silicate weighing 0.8000 gram, a mixture 
of NaCl and KC1 weighing 0.2400 gram was obtained. The chlorides were 
dissolved in water, 50.00 ml. of 0.1000 N AgNO 3 added, and the excess of silver 
titrated with KCNS solution, ferric alum being used as an indicator. In the 
last titration, 14.46 ml. were used, and the reagent was exactly 0.30 per cent 
stronger in normality than the AgNO 3 solution. Find the percentage of K 2 O 
and of Na 2 O in the silicate. 

731. A sample of feldspar contains 7.73 per cent Na 2 O and 9.17 per cent 
K 2 O. What must be the normality of a silver nitrate solution if 25.18 ml. 
precipitates the chloride ions from the combined chlorides in a sample weighing 
0.1500 gram? 

732. A mixture of pure chlorides of potassium, lithium, and sodium weigh- 
ing 0.4800 gram is obtained from 1.600 grams of a silicate. The chlorine in 
this mixture is equivalent to 80.00 nil. of 0.1110 N silver nitrate solution, and 
the potassium is equivalent to 0.1052 gram of K 2 PtCl 6 . Compute the per- 
centage of K 2 O and of Li 2 O in the silicate. 

733. A mixture of LiCl and BaBr 2 weighing 0.5000 gram is treated with 
37.60 ml. of 0.2000 N silver nitrate and the excess of the latter titrated with 



248 CALCULATIONS OF ANALYTICAL CHEMISTRY 

18.50 ml. of 0.1111 N thiocyanate solution. Find the percentage of Ba in the 
mixture. 

734. Express the calculation of the per cent Na 2 O in a silicate containing 
sodium and potassium from the following data: Weight of sample = A grams. 
Weight of NaCl + KC1 obtained = B grams. Weight of AgN0 3 crystals 
added to precipitate the chlorine from these chlorides and give an excess = C 
grams. Volume of D normal KCNS required to titrate the excess silver 
ions = E ml. 



CHAPTER XV 

COMPLEX-ION FORMATION METHODS 
(COMPLEXIMETRY) 

96. Equivalent Weights in Complex-ion Methods. Reactions 
in which complex ions are formed are common in chemistry, par- 
ticularly in qualitative analysis, and many of them should already 
be familiar to the student. Typical cases where complex ions are 
formed are the following: 

Ag+ + 2NH 4 OH - Ag(NH,) a + + 2H 2 

Cd++ + 4CN- - Cd(CN) 4 - 
6C1- 



Unfortunately, because of lack of suitable indicators, few of the 
many reactions of this class can be used as a basis for a volu- 
metric analysis. Most of those that are in common use are cov- 
ered by the examples and problems below. 

The gram-equivalent weight of a substance involved in a com- 
plex-ion forming reaction is based as usual on 1.008 grams of H+ 
as the standard of reference. As in all previous cases, if one for- 
mula weight of a substance reacts with A hydrogen equivalents, 
its equivalent weight is its formula weight divided by A. The 
milliequivalent weights of the metal ions in the above four equa- 
tions are Ag/2,000, Cd/4,000, Sn/6,000, and Hg/4,000, respect- 
ively. Conversely, the equivalent weight of CN~ in the second 
reaction is 4CN/2,000 = CN/500, each atom of Cd being con- 
sidered as equivalent to 2 atoms of H+ . 

EXAMPLE I. LIEBIG METHOD. How many grams of NaCN 
are present in a solution that is titrated just to a permanent 
turbidity with 26.05 ml. of AgNO 3 solution containing 8.125 grams 
of AgNO 3 per liter? 
SOLUTION: 

2CN- + Ag+- 
249 



250 CALCULATIONS OF ANALYTICAL CHEMISTRY 

The next drop of AgNOs gives a permanent precipitate of Ag2(CN) 2 , 
which serves as the indicator for the above reaction. 

Ag(CN) 2 ~ + Ag+ -> Ag 2 (CN) 2 (indicator) 

812^ 
Normality of AgN0 3 = - 0.04782 N 



The milliequivalent weight of NaCN in this case is not NaCN/- 
1,000 since two cyanide ions react with one silver ion in the titra- 
tion. The milliequivalent weight of NaCN may be considered 
here to be 2NaCN/l,000. 
Then, 

Grams NaCN = 26.05 X 0.04782 X 2 f^ N = 0.1221 gram. Ans. 



Or, from another point of view, if 26.05 ml. are necessary to form 
the complex ion as shown above, then twice that amount is neces- 
sary to precipitate the cyanide completely. 

2CN- +'2Ag+ -> A g2 (CN) 2 
Therefore, 

NnPN 

Grams NaCN = 2 X 26.05 X 0.04782 X = 0. 1221 gram. Ans. 



EXAMPLE II. A solution contains KCN and KC1. It is titrated 
with 0.1000 N AgNOs to a faint turbidity, requiring 15.00 ml. 
Then 32.10 ml. more of the AgNOa are added and the precipitates of 
Ag 2 (CN) 2 and AgCl are filtered off. The filtrate requires 7.20 ml. of 
0.08333 N KCNS to give a red color with ferric indicator. How 
many grams of KCN and of KC1 are present in the original solution? 
SOLUTION: 



TCPN 

15.00 X 0.1000 X ^-^ = 0.1953 gram KCN. Ans. 
oUU 

Total volume AgNO 3 added = 15.00 + 32.10 = 47.10 ml. 

AgNOs required to precipitate KCN completely as 

Ag2(CN) 2 = 2 X 15.00 = 30.00 ml. 

47.10 - 30.00, - 17.10 ml. AgNO 3 (reacting with KC1 and giving 

excess) 

(17.10 X 0.1000) - (7.20 X 0.08333) X 



0.08276 gram KC1. Ans. 



COMPLEX-ION FORMATION METHODS 251 

EXAMPLE III. VOLUMETRIC NICKEL. How many grams of Ni 
are contained in an ammoniacal solution that is treated with 
49.80 ml. of KCN solution (0.007810 gram per milliliter) and the 
excess KCN titrated with 5.91 ml. of 0.1000 N AgN0 3 , KI being 
used as an indicator? 

SOLUTION: The essential reactions are 



6 +H - + 4CN- + 6H 2 O - Ni(CN)r + 6NH 4 OH 
2CN- + Ag+-Ag(CN) 2 - 

Unlike the Liebig method above, the formation of Ag2(CN)2 can- 
not be used as an indicator since this salt is soluble in NH 4 OH. 
Instead, excess Ag+ is indicated by the formation of Agl which 
is insoluble in NH 4 OH. 



Normality KCN solution = = 0.1200 N 

IvUJN / 1,UUU 



5.91 ml. AgNO 3 =0= 5.91 X X 2 = 9.85 ml. KCN solution 



(see Example I) 
Net milliliters of KCN = 49.80 - 9.85 = 39.95 ml. 

39.95 X 0.1200 X - = 0.07032 gram Ni. Ans. 



EXAMPLE IV. A volumetric method for zinc consists in titrat- 
ing it in acid solution with a standard solution of K4Fe(CN)e. 
The reaction takes place in two steps, the net reaction being 

3Zn++ + 2Fe(CN) 6 s + 2K+ -> K 2 Zn 3 [Fe(CN) 6 ]2 

Ferric ions or uranyl ions are used to indicate the completion of 
the reaction (by forming a highly colored insoluble ferrocyanide). 
If 15.5 ml. of a solution of K4Fe(CN) 6 which is tenth-normal as 
a potassium salt is used in a given titration, what weight of zinc 
is shown to be present? 
SOLUTION: 

1 F.W. K 4 Fe(CN) 6 as a salt = 4 gm.-atoms H+ 
Therefore each gm.-atom Zn++ =0= % gm.-atoms H+ 

15.5 X 0.100 X ^^ = grams Zn 

= 0.380 gram. Ans. 



252 CALCULATIONS OF ANALYTICAL CHEMISTRY 

Problems 

735. How many milliliters of 0.1000 N AgNOa are required to titrate to a 
faint permanent turbidity a solution containing 10.00 millimoles of KCN? 

Ans. 50.00 ml. 

736. A solution containing KCN and KC1 requires 20.0 ml. of 0.100 N 
AgNO 3 solution to titrate the KCN to a faint turbidity by the Liebig method. 
After addition of 50.0 ml. more of the silver solution and filtering, the filtrate 
requires 16.0 ml. of 0.125 N KCNS, ferric alum being used as an indicator. 
Calculate the number of millimoles of KCN and of KC1 in the original 
solution. 

Ans. 4 millimoles KCN, 1 millimole KC1. 

737. A sample consists of 80.00 per cent KCN, 15.00 per cent KC1, and 
5.00 per cent K 2 SO 4 . A half-gram sample would require how many milliliters 
of 0.1000 molar AgNOa for titration to a faint permanent turbidity? If 
80.00 ml. more of the AgNOa were added, how many milliliters of 0.2000 molar 
KCNS would be required to complete the titration? 

Ans. 30.71 ml. 19.62 ml. 

738. A powder containing KCN, KCNS, and inert material weighs 1.200 
grams, and the solution of it requires 23.81 ml. of 0.08333 N AgNOs to titrate 
the KCN by the Liebig method. A 50-ml. pipetful of the silver solution is 
then added, and the precipitated AgCN and AgCNS are filtered. The filtrate 
requires 10.12 ml. of 0.09090 KCNS for the excess silver, ferric ions being 
used as the indicator. Calculate the percentage of KCN and of KCNS in the 
powder. 

Ans. 21.53 per cent KCN, 10.21 per cent KCNS. 

739. Zinc can be determined by direct titration with standard K 4 Fe(CN)6 
and the net reaction is as follows: 3ZnCl 2 + 2K4Fe(CN) 6 - K 2 Zn 8 [Fe(CN) 6 ]2 + 
6KC1. If the KiFe(CN)6 is 0.1000 N as a potassium salt, what is the value 
of each milliliter of it in terms of grams of Zn? If the KiFe(CN)6 were 
0.1000 N as a reducing agent (in reactions where it is oxidized to ferricyanide), 
what would be the value of 1 ml. of it in terms of zinc? 

Ans. 0.002452 gram. 0.009807 gram. 

740. Find the weights of dissolved KC1, KCN, and KCNS in 500 ml. of a 
solution that analyzed as follows: 30.0 ml. of the solution titrated for KCN 
by the Liebig method reacted with 9.57 ml. of AgNO 3 solution (15.0 grams per 
liter). Then 75.0 ml. more of the silver solution were added, and the solution 
was filtered. The filtrate contained enough silver to react with 9.50 ml. of 
0.100 N KCNS. The precipitate was heated with HNO 8 to decompose the 
AgCN and AgCNS, the H 2 S0 4 formed was precipitated with barium nitrate, 
and the solution then reacted with 58.4 ml. of 0.100 N KCNS. 

Ans. KC1 - 0.85 gram, KCN = 1.83 grams, KCNS 6.73 grams. 



COMPLEX-ION FORMATION METHODS 253 

741. A solution containing ^ millimole of KC1, J millimole of KCN, and 
Y millimole of KCNS is titrated with 0.0667 M AgNO 8 to a faint turbidity, 
requiring A ml. Then enough more of the AgNO 3 is added to make a total of 
30.00 ml. of the AgNO 3 . The precipitate is filtered off and the filtrate requires 
B ml. of 0.100 M KCNS to give a red color with ferric alum indicator. The 
precipitate is decomposed with concentrated HNO 3 and the solution is diluted 
which leaves only AgCl as a residue. The nitric acid solution containing the 
silver from the Ag 2 (CN) 2 and the AgCNS is titrated with 0.100 M KCNS, 
requiring C ml. What are the values of A, B, and C? 

Ans. A = 2.50, B = 9.17, C = 5.83. 

742. A nickel ore contains 10.11 per cent Ni. A half-gram sample is decom- 
posed and the ammoniacal solution treated with 60.00 ml. of a 0.08333 M 
solution of KCN. A little KI is added as an indicator and the solution is 
titrated with 0.06667 M AgNO 3 to a faint permanent turbidity. What volume 
of the AgNO 3 is required? 

Ans. 11.65 ml. 

743. Find the percentage of nickel in an ore if the sample weighs 0.3000 
gram and the ammoniacal solution is treated with 20.00 ml. of KCN (31.2 
grams per liter) and then requires 14.00 ml. of AgNOs (25.5 grams per liter), 
KI being used as an indicator. 

Ans. 26.4 per cent. 



744. What weight of KCN is equivalent to 30.00 ml. of AgNO 3 solution con- 
taining 15.00 grams per liter (a) by the Volhard method for cyanide and (&) by 
the Liebig method? 

746. A sample of impure KCN weighs 0.950 gram and requires 22.0 ml. 
of 0.0909 N AgNO 3 to obtain a turbidity in the Liebig titration. What is the 
percentage of KCN? If the sample contained also 0.102 gram of NaCl, what 
additional volume of AgNO 3 would be required for complete precipitation? 

746. A sample containing KCN weighs 1.000 gram and requires 24.00 ml. 
of 0.08333 N AgNO 3 solution to obtain a faint permanent turbidity. What is 
the percentage of KCN? If the sample also contained 10.00 per cent KC1, 
what volume of the AgN0 3 solution would be required to precipitate the KCN 
and KC1 completely? 

747. A solution is known to contain dissolved KC1, KCNS, and KCN. 
The solution is titrated to a faint turbidity by the Liebig method for cyanide 
with 25.00 ml. of 0.0880 N AgNO 3 solution. A 100-ml. pipetful of the AgN0 3 
is then added and the solution is filtered; the excess silver in the filtrate requires 
50.4 ml. of 0.0833 N KCNS solution. The precipitate of the three silver salts 
is boiled with HNO 3 , which decomposes the AgCN and AgCNS and leaves 
the AgCl, which is filtered off. The filtrate requires 65.0 ml. of the above- 
mentioned KCNS solution of the silver. Calculate the number of milligrams 
of KCN, KC1, and KCNS in the original solution. 



254 CALCULATIONS OF ANALYTICAL CHEMISTRY 

748. A mixture of KCNS, KCN, KC1 weighing 0.687 gram reacts with 
30.0 ml. of 0.0500 N AgNO 3 in the Liebig titration and with 150 ml. more in the 
Volhard titration. Find the percentage composition of the original powder. 

749. What is the percentage of nickel in an ore if the ammoniacal solution 
of a 1.000-gram sample is treated with 3.255 grams (= 50 millimoles) of KCN 
and the excess KCN requires 50.00 ml. of 0.1000 molar AgNO 3 to obtain a 
turbidity with KI indicator? 

750. The Ni in a 0.9000-gram sample of millerite is converted to the 
ammonia complex, and to the solution are added 0.25 ml. of AgNO 3 solution, 
containing 20.00 grams AgNO 3 per liter, and 5.00 ml. of a KI solution that 
serves as the indicator. By addition of two 10-ml. pipetfuls of KCN solution 
(13.00 grams KCN per liter) the turbidity due to the Agl is found to have 
disappeared, but it just reappears on addition of exactly 1.50 ml. more of the 
AgN0 3 . Calculate the percentage of Ni in the millerite. 

761. How many grams of copper are represented by each milliliter of KCN 
in the cyanide method for determining copper (see Part VI, under Copper), if 
each milliliter of the KCN is equivalent to 0.01000 gram of silver by the Liebig 
method? 



PART IV 
ELECTROMETRIC METHODS 

CHAPTER XVI 
POTENTIOMETRIC TITRATIONS 

97. Potentiometric Acidimetric Titrations. In a potentiometric 
titration the principles discussed in Chap. VI are applied in a 
practical way. Suppose a solution of hydrochloric acid is to be 
titrated potentiometrically with a standard solution of sodium 
hydroxide. One method is to use a hydrogen electrode (con- 
sisting of a platinum electrode coated with platinum black and 
over which pure hydrogen gas is allowed to bubble) immersed in 
the solution. This is one half cell. The other half cell is a calomel 
cell. This consists of a tube containing free mercury in contact 
with a solution saturated with mercurous chloride and usually 
either one molar with respect to chloride or saturated with po- 
tassium chloride. The two half cells are connected by means of 
a capillary tube filled with potassium chloride solution. The whole 
cell (assuming one-molar chloride to be used) is expressed as 
follows: 



Hg | Cl- (1 molar), MHg 2 Cl 2 1| H+, i^H 2 (1 atmosphere) | Ft 

At 25 the electrode potential of this so-called normal calomel cell 
is +0.285 volt (see Table XI, Appendix). Using saturated KC1, 
the potential is +0.246 volt. 

(1) Hg + Cl- = ^Hg 2 Cl 2 + e E l = +0.285 

(2) HH 2 (1 atmosphere) = H+ + e 

F i7o . 0.0591, [H+] 

^2 = /V H -- i log 7 - ~^ u 
1 (press. H 2 ) ! /2 

= + 0.0591 log [H+] 
E = Ei-Ei = 0.285 - 0.0591 log [H+] 



255 



256 CALCULATIONS OF ANALYTICAL CHEMISTRY 

By measurement of the e.m.f . of the cell, the pH value of the solu- 
tion can be determined from this formula. Furthermore, the pH 
values of the solution can be determined in the same way at suc- 
cessive points in the titration and those pH values plotted against 
corresponding buret readings. There is obtained a curve similar 
to curve (A) (A) in Fig. 3, Sec. 82. If the nearly vertical line of 
inflection is bisected, the volume of titrating solution correspond- 
ing to the equivalence point can be read off. Since the titration 
is independent of color indicators, the titration can be as suc- 
cessfully carried out in a dark-colored or turbid solution as in a 
colorless one. Plotting the results of potentiometric titrations of 
weak acids like acetic acid and weak bases like ammonia gives 
curves like (C) (C) and (D) (D) in Fig. 3, and the pH value at 
the equivalence point or at any other stage of the titration can 
be readily found in each case. 

EXAMPLE. If, at the equivalence point in the titration of a 
certain solution of acetic acid, pH = 9.10, what e.m.f. should be 
given by the cell made up of this solution in contact with a hy- 
drogen electrode and a normal calomel half cell? 
SOLUTION : 

E - 0.285 
P 0.0591 

9.10X0.0591 -#-0.285 

E = 0.823 volt. Ans. 

98. Simple Potentiometric Titration Apparatus. The essential 
parts of a potentiometric titration apparatus of the type discussed 
above are shown in diagrammatic form in Fig. 7. An outer cir- 
cuit consists of a storage battery S, a rheostat 72, and a slide wire 
M 0, of uniform diameter. The inner circuit consists of a sensi- 
tive galvanometer G, a key K for closing the circuit, and the cell 
to be measured. The direction of the current in the inner circuit 
opposes that in the outer circuit so that, when the position of N 
on the slide wire is adjusted so that the two voltages are equal, 
no current will flow through the galvanometer. In determining 
the voltage of the cell to be measured, the position of N on the 
slide wire is adjusted until the galvanometer needle no longer de- 
flects when the key is momentarily closed. The distance MN 
along the slide wire is then a measure of the desired voltage. If 



POTENTIOMETRIC TITRATIONS 



257 



R 



M 



vwwvyww 



N 




a standard Weston cell of known voltage is previously inserted in 
place of the cell to be measured, the resistance at R can be so ad- 
justed that the scale divisions beside the wire will directly register 
millivolts or some simple 
multiple thereof. 

As a matter of fact, 
the purpose of a potentio- 
metric titration is usually 
to establish the buret 
reading at the equiva- 
lence point rather than to 
determine voltages or pll 
values with a high degree 
of precision. Since the 
buret reading at the 
equivalence point can be 
established from the mid- 
point of the inflection of 
the curve obtained by 
plotting either e.m.f. or 



FIG. 7. Potentiometric titralion hook-up. 



pH values against corresponding buret readings, the use of a simpler 
apparatus than the one just described is possible. In this apparatus, 
a simple dry cell is substituted for the storage battery, and a simple 
voltmeter is used in place of the slide-wire arrangement. After the 
addition of each increment of titrating solution, the resistance at R 
is adjusted until, on closing the key, the galvanometer needle does 
not deflect. The voltmeter is then read directly. When these 
voltages are plotted against buret readings, a curve is obtained 
that inflects sharply at the equivalence point. 

99. Quinhydrone Electrode. Several substitutes for the hydro- 
gen electrode are available, their principal advantages being the 
elimination of the cumbersome purifying trains necessary for the 
hydrogen electrode. Among these substitutes is the quinhydrone 
electrode. This consists of a few crystals of quinhydrone added 
directly to the solution to be titrated. A plain platinum wire 
serves as the metallic contact, and a calomel cell is used as the 
other half cell. 

When quinhydrone is added to water, a very small amount dis- 
solves and dissociates into an equimolecular mixture of quinone 



258 CALCULATIONS OF ANALYTICAL CHEMISTRY 



(C 6 H4O 2 ) and hydroquinone (CeEUC^Hb). These two substances 
are in equilibrium with each other, as shown by the equation 

C 6 H 4 2 + 2H+ + 2e 



The potential of this electrode is a function of the hydrogen-ion 
concentration. 

00591 
, , -j- log 



= 0.700 -1- 0.0591 log [H+] at 25C. 

Using a calomel cell as the other half cell and making it the nega- 
tive electrode in the outer circuit (positive to quinhydrone in the 
inner circuit) we have 

E l = +0.700 + 0.0591 log [H+] 
E 2 = +0.285 

E = E!-E 2 = 0.415 + 0.0591 log [H+] 
or 

TT , rrr+T 0.415 - E 

pH = - log [H+] = - 



In a titration, the quinhydrone electrode being used, the value 
of E becomes zero at about pH = 7; and on the alkaline side of 
this point the calomel cell is used as the positive electrode, and 
the values of E are given a negative sign. Correct values are not 
obtained in solutions where pH > 9. 

100. Glass Electrode. The glass electrode consists of a thin- 
walled bulb of special glass containing an electrode immersed in 
a standard reference solution (e.g., a platinum wire in a quinhy- 
drone-hydrochloric acid solution). The glass bulb serves as a 
semipermeable membrane between the reference solution and the 
solution to be tested. Because of the high resistance of the bulb, 
a galvanometer of high sensitivity or an electrometer potentiom- 
eter is required to measure pH. In some forms the glass elec- 
trode is rather fragile, but this disadvantage is more than offset 
by the facts that (1) there is no contamination of the solution 
being tested, (2) measurements can be made on very small volumes 
of solution, and (3) the presence of oxidizing or reducing agents 
does not influence the results of a pH determination. 



POTENTIOMETRIC TITRATIONS 



259 



Because of variations in composition of the glass used in the 
glass electrode, the formula for determining the pH value with a 
given electrode is usually provided by the manufacturer, but 
slight changes can occur over long periods of time owing to crystal- 
lization of the glass. 

Most modern portable pH meters use glass electrodes and calo- 
mel cells in compact form and are of such construction that pH 
values can be read directly from the instrument. By means of 
such a meter a pll measurement can be made very quickly, and 
the manipulative technique involved is hardly more than that of 
turning a knob a^d pressing a button. 

101. Potentiometric Redox Titrations. The hookup for the 
potentiometric titration of a reducing or oxidizing agent is similar 
to that of an acidimetric titration except that a plain platinum 
wire serves as the electrode. 

Suppose a solution of ferrous sulfate is titrated with a standard 
solution of eerie sulfate (Fe++ + Ce+ ++ + - Fe+++ + Ce+++). Be- 
fore the equivalence point is reached, the principal equilibrium is 
that between ferrous and ferric ions (Fe + + = Fe+++ + e) and the 
ratio of the two concentrations changes rapidly. During this part 
of the titration the cell is represented by 



Pt 



Fe 



HHg 2 Cl 2 , Cl- (1 molar) 



Hg 



fTTp-l-K-1 

= +0.748 + 0.0591 log y J 
E z = +0.285 (calomel cell) 
E = E!-E 2 = 0.463 + 0.0591 log 



Beyond the equivalence point the predominating equilibrium is 
that between cerous and eerie ions (Ce" 1 " 1 " 1 " = Ce ++++ + ). Dur- 
ing this part of the titration the cell is represented by 



Pt 



Ce 
Ce 



HHg 2 Cl 2 , Cl- (1 molar) 



Hg 



260 



CALCULATIONS OF ANALYTICAL CHEMISTRY 



., , 0.0591 , [Ce^ j 
El o + __ log i___ 



= +1.45 + 0.0591 log [Ce +++j 
E 2 = +0.285 (calomel cell). 
E - E l - E 2 - 1.16 + 0.0591 log 



The graph of a typical titration of this type is shown hi Fig. 8. 
When dichromate is used for the titration of iron, the graph be- 
yond the equivalence point depends on the hydrogen-ion concentra- 
tion of the solution since in this case the predominating equilibrium 



s 



+ 7H 2 = Cr 2 7 = + 14H+ + Ge, and 



. 30 



0.0591 lo [Cr 2 7 ~][H+] 14 



6 



[Cr+++] 2 



In all titrations of this kind when the potentials are plotted 
against volumes of titrating solution added, the equivalence point 

is found by bisecting 
the nearly vertical part 
of the curve. As a 
matter 
change 



+1.50 



+1.40 



+1.30 



+1.20 



+1.10 



E 



+1.00 



+0.90 



+0.80 



+0.70 
+0.60 




fact, the 
in e.m.f. is 
usually so great that it 
is often unnecessary to 
tabulate the values for 
the e.m.f. in order to 
determine the volume 
of titrating solution 
corresponding to the 
equivalence point. The 
titrating solution is 
added in small incre- 
ments until the volt- 
meter shows a very 
sudden deflection, and 

the volume is read directly from ^he buret. 

Potentiometric titrations can, of course, be applied to oxidation 

reactions other than the change from ferrous to ferric ions. The 



Titration of 30 ml. FeS0 4 
(N/10) with N/10 Ce <S0 4 ) 2 



Of 



10 20 30 



40 
ML 



50 60 70 80 



FIG. 8. 



POTENTIOMETRIC TITRATIONS 261 

e.m.f . at the equivalence point is different for different reactions, 
but the sudden change in voltage is common to all. 

102. Potentiometric Precipitation Titrations. The potentio- 
metric principle can be applied to certain precipitation titrations. 
For example, in the titration of silver ions with halide ions, the 
concentration of silver ions changes during the progress of the 
titration. Using a silver electrode and a regular calomel half cell, 
we have the cell 



Ag Ag+ I MHg 2 Cl 2 , Cl- (1 molar) 1 Hg 

o n^Qi 
E l = +0.799 + ^p log [Ag+] 

#2 - +0.285 (calomel cell) 
E = Ei - # 2 = +0.514 + 0.0591 log [Ag+] 

The graph of a titration of this type shows a sudden inflection at 
the equivalence point as in the case of acidimetric titrations and 
oxidation titrations. 

Problems 

762. A cell made up of a certain basic solution and normal hydrogen- 
calomel electrodes gives at 25C. an e.m.f. of 0.825 volt. What is the pH 
value and what is the hydroxyl-ion concentration of the solution? 

Ans. 9.14, 1.38 X 10~ 6 molar. 

763. A certain solution of sulfuric acid has a hydrogen-ion concentration 
of 3.60 X 10~ 3 . What is the pOH value? With regular hydrogen-calomel 
electrodes, what e.m.f. could be obtained at 25C.? 

Ans. 11.56. 0.429 volt. 

764. With quinhydrone-calomel electrodes, approximately what e.m.f. 
could be obtained at 25C. with a 0.0500 N solution of acetic acid (ionization 
constant 1.86 X 10~ 5 )? What e.m.f. could be obtained from the same solution 
containing an additional mole of acetate ion per 500 ml.? 

Ans. 0.237 volt. 0.041 volt. 

766. With quinhydrone-calomel electrodes, a tenth-molar solution of a 
certain monobasic acid at 25C. gives an e.m.f. of 275 mv. What is its 
approximate ionization constant? 

Ans. 1.83 X 10~ 4 . 

766. What is the hydroxyl-ion concentration of a solution which at 25C. 
and with quinhydrone-calomel electrodes gives an^e.m.f. of zero? 
Ans. 1.07 X 10- 7 . 



262 CALCULATIONS OF ANALYTICAL CHEMISTRY 

757. Calculate the potential at 25C. obtainable from the cell made by 
connecting the half cell: 

Sn++ (0.0600 molar) 
Sn++++ (0.00100 molar) 
with a calomel half cell. 
Ans. 0.208 volt. 



Pt 



768. Plot the foll&wing values of millivolts against milliliters of 0.100 
N NaOH in the potentiometric titration of 2.50 grams of vinegar, hydrogen- 
calomel electrodes being used, and calculate the percentage of acetic acid in 
the vinegar. At what volume of NaOH is the solution neutral, and what 
volume corresponds to the equivalence point? What is the pll value at the 
equivalence point? 0.0 ml. = 420 mv.; 4.0 ml. = 475; 8.0 ml. = 540; 
12.0 ml. = 588; 16.0 ml. = 620; 18.0 ml. = 638; 19.0 ml. = 650; 19.4 ml. = 
670; 19.8 ml. = 790; 20.0 ml. = 830; 20.2 ml. = 856; 20.5 ml. = 875; 
21.0 ml. = 900; 22.0 ml. = 930; 24.0 ml. = 948; 28.0 ml. = 970; 32.0 ml. = 
985. 



55. 
Ans. 4.75 per cent. 19.2 ml., 19.8 ml. 8.55. 



769. A sample of sodium carbonate containing inert impurities weighs 
1.10 grams. It is dissolved in water and titrated potentiometrically with 
0.500 N HC1, normal hydrogen-calomel electrodes being used. Plot the 
following values of milliliters against corresponding millivolts, and calculate 
the approximate percentage of Na 2 CO 3 in the sample. What are the 
pH values at the two equivalence points? 0.01 ml. = 928 mv.; 5.0 ml. = 922; 
10.0 ml. = 912; 12.5 ml. = 900; 15.0 ml. = 880; 17.5 ml. = 838; 20.0 ml. = 
762; 22.5 ml. = 710; 25.0 ml. = 696; 27.5 ml. = 682; 30.0 ml. = 669; 
32.5 ml. = 650; 35.0 ml. = 607; 37.5 ml. = 484; 40.0 ml. = 452; 45.0 ml. = 
427; 50.0 ml. = 416. 

Ans. 87 per cent. 9.1, 4.5. 

760. A sample of Na 2 C0 3 is known to contain either NaOH or NaHCO 3 , 
together with inert matter. A sample weighing 1.50 grams is titrated potentio- 
metrically with 0.600 N HC1. Plot milliliters of acid against millivolts, and 
determine the approximate percentage composition of the sample. 0.1 ml. = 
930 mv.; 5.0 ml. = 918; 10.0 ml. = 899; 12.5 ml. = 872; 15.0 ml. = 820; 
17.5 ml. = 757; 20.0 ml. = 727; 22.5 ml. = 708; 25.0 ml. = 696; 27.5 ml. = 
683; 30.0 ml. = 668; 32.5 ml. = 648; 35.0 ml. = 606; 37.5 ml. = 485; 40.0 
ml. = 452; 45.0 ml. = 427; 50.0 ml. = 416. 

Ans. 63.7 per cent Na 2 CO 3 , 20.1 per cent NaHC0 3 . 

761. A sample of formic acid (IICOOTI) is dissolved in water and titrated 
potentiometrically with 0.400 N NaOH, quinhydrone-calomel electrodes 
being used. Plot the titration curve from the following data, and calculate 
the pH value at the equivalence point. Approximately how many grams of 
HCOOH are shown to be present in the solution? 0.0 ml. = 273 mv.; 
10.0 ml. = 262; 25.0 ml. = 242; 35.0 ml. = 225; 45.0 ml. = 195; 55.0 ml. = 



POTENTIOMETRIC TIT RAT IONS 263 

135; 60.0 ml. = 58; 62.5 ml. = 0; 65.0 ml. = -100; 70.0 mi. = -223; 
80.0ml. = -308. 

Ans. 8.11. 1.2 grams. 

762. In the potentiometric titration at 25C. of 50.0 ml. of 0.100 N ferrous 
sulfate (diluted to 250 ml.) with 0.100 N eerie sulfate, what should be the 
voltage reading (a) at the equivalence point, (6) halfway to the equivalence 
point? 

Ans. (a) 1.10 volts, (b) 0.748 volt. 

763. A cell is made up of a platinum wire, dipping into a solution of cerous 
and eerie ions, and a regular calomel cell. At 25C. an e.m.f. of 1,190 mv. 
is obtainable. Calculate the ratio of concentration of eerie ions to concentra- 
tion of cerous ions in the solution. 

Ans. 3.22. 

764. A sample of limonite weighing 0.350 gram is dissolved in HC1, and 
the ferric ions are reduced by means of a slight excess of stannous chloride. 
Without removal of the excess stannous ions the solution is titrated poten- 
tiometrically with 0.100 N eerie sulfate solution, platinum-calomel electrodes 
being used. Plot the following values of milliliters of eerie sulfate against 
corresponding millivolts, and from the graph calculate the approximate 
percentage of Fe 2 O 3 in the sample. The stannous ions are oxidized by the 
eerie sulfate first. 0.0 ml. = 190 mv.; 1.00 nil. = 218; 2.00 ml. = 223; 
3.00 ml. = 240; 4.00 ml. = 325; 5.00 ml. = 342; 6.00 ml. = 350; 9.00 ml. = 
363; 15.0 ml. = 382; 20.0 ml. = 388; 25.0 ml. = 393; 30.0 ml. = 417; 32.0 
ml. = 450; 34.0 ml. = 510; 35.0 ml. = 570; 36.0 ml. = 910; 37.0 ml. = 
1,100; 39.0 ml. =1,155; 45.0 ml. = 1,217; 50.0 ml. = 1,229. 

Ans. 73 per cent. 

766. Twenty-five milliliters of 0.200 N AgN0 3 are diluted to 250 ml. and 
titrated potentiometrically with 0.200 N KBr, a silver electrode and a normal 
calomel electrode being used. Assuming the solubility of silver bromide to 
be 5.9 X 10~ 7 mole per liter, calculate the theoretical value for E (a) when a 
fraction of a drop of bromide has been added, (fe) at the equivalence point, 
(c) after 20 ml. of the bromide have been added. (Hint: Find the silver-ion 
concentration in each case, and use the specific oxidation potential of Ag = 
Ag+ + e given in the Appendix.) 

Ans. (a) +0.414 volt, (b) +0.146 volt, (c) -0.039 volt. 



766. A certain solution of sodium hydroxide has a hydroxyl-ion concen- 
tration of 5.20 X 10" 4 . With regular hydrogen-calomel electrodes, what 
e.m.f. could be obtained at 25C.? 

767. With quinhydrone-calomel electrodes, a certain solution at 25C. 
gives an e.m.f. of 116 mv. What is the hydroxyl-ion concentration of the 
solution? 

768. With regular hydrogen-calomel electrodes, a hundredth-molar solution 
of a certain monoacidic base gives at 25C. an "e.m.f. of 946 mv. What is the 



264 CALCULATIONS OF ANALYTICAL CHEMISTRY 

approximate ionization constant of the base? What e.m.f. would be obtained 
if 2.00 moles per liter of cation common to the base were introduced into the 
solution? 

769. Plot values on graph paper showing the relationship between milli- 
volts and pH values (a) when using normal hydrogen-calomel electrodes, 
(6) when using quinhydrone-calomel electrodes. Include the range between 
pH = 14 and pH = - 2. 

770. Prove that in the potentiometric titration of ferrous ions with eerie 
ions the voltage at the equivalence point is given by the general expression 
Ei = (Ei + -E 2 )/2. (Hint: At the equivalence point, not only is the reaction 
Fe^ + Ce+ ++ + = Fe+++ + Ce ++ + at equilibrium, and hence E l = # 2 , but 
also [Fe++] = [Ce+++ + ] and [Fe+ ++ ] = [Ce ++ +].) 

771. In the potentiometric titration of 60.0 ml. of 0.100 N ferrous sulfate 
(diluted with water) with 0.200 N eerie sulfate, what should be the voltage 
reading (a) after 10.0 ml. of eerie sulfate has been added, (6) after 30.0 ml. of 
eerie sulfate has been added? 

772. The potentiometric titration of a 25-ml. pipetful of 0.268 N H 2 S04 
with NaOH solution gave the following values of millivolts for the corre- 
sponding volumes of NaOH: 0.0 ml. = 369 mv.; 5.0 ml. = 378; 10.0 ml. = 
388; 15.0 ml. = 398; 20.0 ml. = 406; 25.0 ml. - 420; 28.0 ml. = 460; 
30.0 ml. - 516; 30.5 ml. = 690; 31.0 ml. = 860; 35.0 ml. = 949; 40.0 ml. = 
966; 45.0 ml. = 982. Plot the millivolts of potential as ordinates against 
milliliters of NaOH as abscissas, and from the curve determine the pH value 
at the equivalence point and the normality of the NaOH solution. 

773. Make a graph for the following potentiometric titration of 40.0 ml. 
of 0.213 N H 3 PO 4 diluted with water to 200 ml. and titrated with 0.200 N 
NaOH at 25C., hydrogen-calomel electrodes being used. Calculate the 
pH value at which the replacement of the first and second hydrogens of 
H,PO 4 occurs. 0.0 ml. = 300 mv.; 5.0 ml. = 315; 10.0 ml. =* 350; 13.0 ml. = 
385; 13.5 ml. = 398; 13.8 ml. = 405; 14.0 ml. = 415; 14.2 ml. = 450; 14.4 
ml. = 525; 14.8 ml. = 555; 15.5 ml. = 566; 17.0 ml. = 580; 20.2 ml. = 603; 
25.0 ml. = 640; 27.5 ml. = 658; 28.5 ml. = 675; 28.8 ml. = 685; 29.0 ml. = 
740; 29.2 ml. = 760; 29.5 ml. = 795; 30.0 ml. = 815; 31.0 ml. = 835; 
35.0 ml. 870; 40.0 ml. = 890. 

774. A cleaner is known to contain, in addition to inert material, either 
NaOH, Na^COs, or mixtures of these. The potentiometric titration of a 
1.00-gram sample in 100 ml. of water with 0.265 N HC1, regular hydrogen- 
calomel electrodes being used, give the following pH values: 0.0 ml. = 11.70; 
5.50 ml. 11.68; 10.0 ml. = 11.66; 20.0 ml. - 11.60; 30.0 ml. = 11.44; 
45.0 ml - 10.98; 55.0 ml. = 9.75; 60.0 ml. = 8.76; 62.0 ml. = 7.66; 
64.0 ml. = 6.31; 66.0 ml. - 5.70; 68,0 ml. = 5.40; 70.0 ml. = 5.05; 72.0 ml. = 
4.81; 73.0 ml. = 4.10; 74.0 ml. = 2.44; 76.0ml. = 1.94; 78.0 ml. = 1.70; 
85.0 ml. = 1.41; 95.0 ml. = 1.16. 

Interpret the curve, stating which components are present and their 



POTENTIOMETRIC TITRATIONS 265 

approximate percentages. What voltage reading is obtained at the first 
point of inflection? 

775. The chromium in 5.00 grams of steel was oxidized to dichromate and 
then titrated potentiometrically with 0.1039 N ferrous sulfate solution. Plot 
the curve, and compute the percentage of Cr from the following data which 
show volts X 10 against milliliters: 0.0 ml. = 6.50; 5.0 ml. 7.90; 10.0 ml. - 
8.00; 15.0 ml. = 8.10; 20.0 ml. = 8.20; 25.0 ml. = 8.40; 30.0 ml. = 8.60; 
35.0ml. = 8.78; 36.0 ml. * 8.85; 37.0ml. = 8.87; 37.5 ml. = 8.87; 38.0 rnl. = 
8.85; 38.3 ml. = 8.84; 38.4 nil. 5.03; 39.0 ml. = 4.45; 40.0 ml. 4.15; 
45.0 ml. = 390. 



CHAPTER XVII 
CONDUCTOMETRIC TITRATIONS 

103. Conductance. Strong acids, strong bases, and most salts, 
when dissolved in a relatively large volume of water, are prac- 
tically completely dissociated into ions. These ions are capable 
of transporting electricity, and because of them the solutions are 
good conductors of the electric current. The conductance of a 
solution is the reciprocal of its electrical resistance and is meas- 
ured in reciprocal ohms or mhos. 

The specific conductance of a solution is the conductance of a 
cube of the solution of 1-cm. edge. The specific conductance at 
25C. of 0.100 N HC1 is 0.0394 mho; the specific conductance of 
0.0100 N HC1 is 0.00401 mho. 

Equivalent conductance is the conductance of a solution contain- 
ing one gram-equivalent weight of dissolved electrolyte between 
electrodes 1 cm. apart. It is therefore numerically equal to the 
product of the specific conductance of the solution and the num- 
ber of milliliters containing one gram-equivalent weight of electro- 
lyte. Thus the equivalent conductance of 0. 100 N IIC1 is 0.0394 X 
10,000 = 394 mhos; the equivalent conductance of 0.0100 N HC1 
is 0.00401 X 100,000 = 401 mhos. As a solution becomes more 
dilute, its equivalent conductance becomes somewhat greater owing 
to the fact that in more dilute solutions inter-ionic effects are les- 
sened, which gives the apparent effect of increasing the degree of 
ionization of the dissolved substance. 

By extrapolation it is possible to determine the equivalent con- 
ductance of a solution at infinite dilution. For hydrochloric acid 
this value at 25C. is 425.8 reciprocal ohms. This is the theoretical 
conductance that would be given by a "perfect" solution con- 
taining 36.46 grams of HC1 between electrodes 1 cm. apart. 

104. Mobility of Ions. Different kinds of ions have different 
velocities, so that when an electric current is passed through a 
solution, the faster moving ions carry a relatively greater amount 

266 



CONDUCTOMETRIC TITRATIONS 267 

of the current. In the case of very dilute hydrochloric acid, the 
hydrogen ions, moving much faster than the chloride ions, carry 
about 82 per cent of the current; the chloride ions carry only 
about 18 per cent. The mobility of an ion is the equivalent con- 
ductance of that ion, and the equivalent conductance of an elec- 
trolyte is equal to the sum of the mobilities of its ions. Thus, 
the equivalent conductance at 25C. of hydrochloric acid at in- 
finite dilution (= 425.8) is equal to the sum of the mobility of 
the hydrogen ions (= 350) and the mobility of the chloride ions 
(= 75.8) at that temperature. If several electrolytes are present 
in a solution, all the ions contribute to the conductance of the 
solution. Mobilities increase by about 2 per cent for each degree 
centigrade increase in temperature. 

Table III gives the equivalent conductances, or mobilities, at 
25C. of some of the common ions at infinite dilution. From it 
can be calculated the equivalent conductances of corresponding 
electrolytes at infinite dilution. 

TABLE III. IONIC CONDUCTANCES OR MOBILITIES 
AT INFINITE DILUTION, 25C. 

Na+ 50.8 Cl~ 75.8 

K + 74.8 Br~.* 77.7 

Ag+ 63.4 I- 76.0 

H+ 350 OTI- 193 

NH 4 + 74.9 C 2 Il3O 2 - 40.8 

41.7 l / 2 SOr 80.0 

55.0 C1O 3 - G3.3 

65.2 NO 3 - 70.9 

61 Br0 3 - 55.3 

71.0 I0r 39.6 

53.6 J^CjOr 73.5 

54 HFe(CN) 6 - 97.3 

68.4 MFe(CN) 6 B 100.8 

105. Conductometric Acidimetric Titrationsl Consider the ti- 
tration of a dilute solution of HC1 with NaOH solution: 

H+C1- + (Na+OH-) - Na+Cl- + H 2 O 

At the beginning of the titration, the HC1 solution has a high 
conductance value, owing principally to the extremely high mo- 
bility of the hydrogen ions. As NaOH is added, the concentra- 
tion of the hydrogen ions is decreased and, although hydrogen 
ions are replaced by sodium ions, the mobility of the latter is 



268 



CALCULATIONS OF ANALYTICAL CHEMISTRY 



much less, so that the conductance of the solution decreases 
rapidly. At the equivalence point, the solution contains only 
NaCl, and the conductance is at a minimum for, on further addi- 
tion of NaOH, the hydroxyl ions with their high mobility give a 
rapidly increasing conductance to the solution. If the titration 



8 



I 




4-" \ 
^ 



,1 






Volume of NaOH 
Fio. 9. Conductometric titration of dilute HC1 with NaOH. 

is carried out under constant conditions of temperature, etc., and 
the volume of titrating solution is plotted against conductance, a 
curve of the appearance of line ABC in Fig. 9. is obtained 

In this figure, line AH represents that part of the conductance 
of the solution contributed by the HC1 alone; DBJ represents 
that part of the conductance of the solution contributed by the 
NaCl alone. Line A B is therefore the resultant of these two curves, 
the distance GI being equal to FI + EL Line HK represents the 
conductance of the excess NaOH alone; line BJ is the conduct- 
ance of the NaCl present in the solution after the equivalence 
point has been reached; and BC is the resultant. 



CONDUCTOMETRIC TITRATIONS 269 

Ideal titration curves applying to perfect solutions can be cal- 
culated from the mobilities of the ions involved. Thus, in the 
titration of a very dilute solution containing a gram-equivalent 
weight of HC1 with a relatively concentrated solution of sodium 
hydroxide (so as to give no appreciable change in the total vol- 
ume of the solution being titrated), the theoretical conductance 
of the original solution is 350 (11+) + 75.8 (Cl~) = 425.8 mhos. 
At the equivalence point the solution contains only NaCl and its 
conductance is 50.8 (Na+) + 75.8 (C1-) = 126.6 mhos. An excess 
of one gram-equivalent weight of NaOH to the resulting solution 
would give a conductance of 50.8 (Na+) + 75.8 (C1-) + 50.8 (Na+) + 
193 (OH") = 370.4 mhos. Plotting these conductance values 
against corresponding relative volumes of NaOH gives a titration 
curve like that of the resultant line ABC in the figure. The 
equivalence point is the intersection of two straight lines. 

In an actual titration of this type the lines are likely to be 
slightly curved because of (1) variation in temperature, due, in 
part at least, to the heat of neutralization, (2) increase in the 
volume of the solution because of added reagent, and (3) inter- 
ionic effects. Foreign ions in the solution may distort the curve 
slightly, although their general effect is" to increase the total con- 
ductance by a constant amount. In spite of this, the inflection 
is sharp and three or four readings on each side of the equivalence 
point are usually sufficient to establish the point of intersection 
and hence the buret reading at the equivalence point. 

The titration of a weak acid like acetic acid with a strong base 
like sodium hydroxide is shown in the curve (a) (a) of Fig. 10. 
Here the first small amount of NaOH will, as before, cause a 
decrease in conductivity but, since the concentration of hydrogen 
ions in acetic acid is small, the conductance of the solution soon 
increases owing to the formation of sodium ions and acetate ions, 
the latter buffering the solution and thus cutting down the con- 
centration of the highly mobile hydrogen ions. The conductance 
values then follow closely those of the sodium acetate formed. 
Beyond the equivalence point the formation of hydroxyl ions does 
not cause a sharp inflection in the titration curve. If, on the other 
hand, NH 4 OH is used to titrate the acetic acid, a curve (6) (&) 
is obtained with a sharper inflection at the equivalence point, for 
the excess NH 4 OH, owing to its slight degree of ionization, has 



270 



CALCULATIONS OF ANALYTICAL CHEMISTRY 



little effect on the conductance of the solution. In an actual titra- 
tion of this type, the two parts of the conductance curve do not 
meet sharply at a point because of hydrolysis effects, but the 
equivalence point can be found by extending the straight parts 
of the titration graph to a common point [see part (c) of Fig. 10]. 




Volume of titrating reagent 

FIG. 10. Conductometric titration of 0.01 N HC 2 H 3 O 2 
(a) with NaOH, (b) with NH 4 OH. 

The titration of a mixture of a strong acid and a weak acid with 
a standard base can often be carried out conductometrically and 
the amount of each acid determined from the graph (see Prob. 
781). A similar type of titration curve is obtained in the titration 
of certain dibasic acids (see Prob. 789). 

106. Conductometric Precipitation Titrations. Many precipi- 
tation titrations are also possible by Conductometric methods. 
Consider, for example, a very dilute solution containing a gram = 
equivalent weight of sodium sulfate being titrated with a con- 
centrated solution of barium acetate. The theoretical conductance 



CONDUCTOMETRIC TITRATIONS 



271 



of the original solution is 50.8 (Na+) + 80.0 (MS0 4 = ) = 130.8 mhos. 
At the equivalence point the conductance is 50.8 (Na 4 *) + 40.8 
(C2H 3 02~) = 91.6 mhos. If an excess of a gram-equivalent weight 
of barium acetate is added, the conductance of the solution is 
50.8 (Na+) + 40.8 (C 2 II 3 O 2 -) + 65.2 (HBa++) + 40.8 (C 2 H 3 O 2 -) - 
197.6 mhos. The titration curve is therefore a flat V-shaped 
one with the equivalence point at the intersection of two straight 
lines. Certain titration curves of this type are illustrated in the 
accompanying problems. 



HH 

E 



"NAAAAA/WWW 



A/ 



1000 -a 



o 



K 



FIG. 11. Conductometric titration hook-up. 

A few complex-forming reactions can also be made the basis 
of conductometric titrations. 

107. Conductometric Titration Apparatus. A simple hookup 
for determining the conductance of a solution is indicated in the 
diagram of Fig. 11, where MO is a slide-wire divided into 1,000 
scale divisions, G is a galvanometer, E is a pair of electrodes so 
fixed as to remain at a constant distance apart and dipping into 
the solution to be measured. S is a source of alternating current, 
stepped down by means of a transformer to about 6 volts, and 
R is a rheostat. 

In measuring relative conductance values in a titration, the 
rheostat is first adjusted so that there is no deflection of the 
galvanometer when the point of contact N is near the center of 
the slide-wire. From then on, the adjustment of the rheostat is 
not changed. The solution is then titrated, and bridge readings are 



272 CALCULATIONS OF ANALYTICAL CHEMISTRY 

taken between each increment of added solution by adjusting the 
slide-wire until no deflection is observed in the galvanometer. 

After any such adjustment, 



a resistance of cell 



1,000 a resistance of rheostat 

but, since the resistance of the rheostat is a constant (K) and the 
resistance of the cell is the reciprocal of its conductance, 

= 1,000 - a 

conductance of cell 
whence, 

o j x * n /% - <A 1 

Conductance of cell = ( -^ 

\ a JK 

Since in an ordinary titration we are not interested in the actual 
conductance values but only in the relative changes in conduc- 
tivity as a means of establishing a titration curve, it is only 
necessary to plot the volume of titrating solution against the values 
(1,000 a) /a as obtained from the bridge readings. Unlike the 
potentiometric graph, the conductometric titration curves are 
straight lines or nearly so, and they can therefore usually be fixed 
by a relatively few volume readings on each side of the equivalence 
point. In order for accurate values to be obtained and in order 
for the lines to be straight or nearly so, it is important to keep 
the temperature of the solution as nearly constant as possible, and 
it is also theoretically necessary that the volume of the solution 
shall not change during the titration. This last condition is ful- 
filled approximately enough for ordinary titrations if the total 
volume of reagent does not exceed 1 or 2 per cent of the solution 
titrated. The reagent should therefore be as concentrated, and 
the solution as dilute, as feasible. A solution 0.01 0.001 normal 
titrated with a 1 normal solution of reagent would be a typical case. 

Problems 

776. If the specific conductance of N/50 IIC1 is 0.00792 mho, what is the 
equivalent conductance of N/50 HC1? 

Ans. 396 mhos. 

777. At 25C. what is the equivalent conductance at infinite dilution of a 
solution of silver sulfate? 

Ans. 143.4 mhos. 



CONDUCTOMETRIC TITRATIONS 273 

778. A solution containing a gram-equivalent weight of BaCla at very high 
dilution is titrated at 25C. with Li 2 SO 4 . From mobilities of the ions involved 
calculate the conductance of the solution (a) at the start of the titration, 
(6) at the equivalence point, and (c) at the point where a total of 2 gram- 
equivalents of Li2SO 4 have been added. Plot these values to show the titration 
graph. Make similar calculations and graph for the titration of BaCl 2 with 
Na2SO 4 . Which gives the sharper inflection at the equivalence point? 

Ans. (a) 141.0, (6) 117.5, (c) 239.2; (a) 141.0, (6) 126.6, (c) 257.4 mhos. 

779. A very dilute solution of sodium hydroxide is titrated conductometri- 
cally with 1.00 N HC1. The following bridge readings of (100 a) /a were 
obtained at the indicated points in the titration. Plot the titration curve and 
from it determine the number of grams of NaOH present in the solution. 
0.00 ml - 3.15; 1.00ml. - 2.60; 2.00 ml. - 2.04; 3.00 ml. - 1.40; 4.00 ml. - 
1.97; 5.00 ml. - 2.86; 6.00 ml. - 3.66. 

Ans. 0.128 gram. 

780. A solution approximately N/100 in sodium acetate is titrated con- 
ductometrically with 1.00 N HC1. From the following titration values showing 
relative conductivities plot the curve and calculate the number of grams of 
NaC2H 3 O2 present in the solution. (Hint: Extend the nearly straight parts of 
the curve to a point of intersection.) 0.00 ml. - 218; 4.00 ml. - 230; 8.00 
ml. - 243; 9.00 ml. - 247; 10.00 ml. - 256; 11.00 ml. - 269; 12.00 ml. - 278; 
14.00 ml. 325; 17.00 ml. 380. Show from mobilities and relative degrees 
of ionization why this form of curve is to be expected. 

Ans. 0.820 gram. 

781. A sample of vinegar has been adulterated with hydrochloric acid. 
It is titrated with 0.500 N NH 4 OH and the following bridge readings of 
(1,000 a)/a were obtained at the indicated buret readings. Calculate the 
number of grams of HC1 and of HC2H 3 O 2 in the sample. (Hint: Find the point 
of neutralization of the HC1 by extending the nearly straight sides of the U-- 
shaped part of the graph to a point of intersection.) 0.00 ml. 2.87; 1.00 ml. 
2.50; 2.00 ml. - 2.10; 2.50 ml. - 1.85; 3.00 ml. - 1.70; 3.20 ml. - 1.70; 
3.50ml. - 1.76; 4.00 ml. - 2.00; 4.20 ml. - 2.10; 4.50 ml. - 2.15; 5.00 ml - 
2.15; 6.00 ml. - 2.14; 7.00 ml. - 2.16; 8.00 ml. - 2.18. 

Ans. 0.0580 gram HC1, 0.0336 gram HdH,Cfc. 



782. At 25C. what is the equivalent conductance at infinite dilution of a 
solution of BaCh? 

783. Sketch the general form of the titration curve you would expect to 
get in the conductometric titration of N/100 NH 4 OII with (a) N/l HC1, 
(6) N/l HC 2 H 8 O2. 

784. Using the eqivalent conductance values obtained from the mobilities 
of the ions, show the general form of the titration curve in each of the following 
cases: (a) titration of NaOH with HNO a , (6) titration of BaCl 2 with K 2 SO 4 , 
(c) titration of BaCl 2 with H 2 SO 4 , (d) titration of Ba(OH) 2 with H 2 SO 4 , 



274 CALCULATIONS OF ANALYTICAL CHEMISTRY 

(e) titration of MgSO 4 with Ba(OH) 2 , (/) titration of NH 4 C1 with NaOH, 
(0) titration of AgNO 8 with LiCl. 

785. In the titration of 80.0 ml. of a solution of HNO 8 with 4.85 N NaOH 
the following relative conductivities were obtained for the corresponding 
volumes of NaOH. Plot the curve and calculate the acid normality of the 
original acid solution. 0.00 ml. - 501; 1.00 ml. 340; 2.00 ml. 175; 
3.00 ml. - 180; 4.00 ml. - 261 ; 5.00 ml. - 338. 

786. A sample of vinegar weighing 5.00 grams is diluted to 500 ml. and ti- 
trated conductometrically with 0.500 N NH 4 OII. The following relative 
conductivities were obtained from the bridge readings of (1,000 a) /a at the 
indicated buret readings: 0.00 ml. - 1.20; 0.50 ml. - 0.95; 1.00 ml. - 0.80; 
1.50 ml. - 0.70; 2.00 ml. - 0.70; 3.00 ml. - 0.95; 4.00 ml. - 1.35; 5.00 
ml. - 1.75; 6.00ml. - 2.13; 7.00ml. - 2.48; 7.50ml. - 2.68; 8.00ml. - 2.78; 
9.00 ml. - 2.82; 10.00 ml. - 2.83; 14.00 ml. - 2.87. Plot these values and 
determine from the graph the buret reading at the equivalence point. From 
this calculate the acidity of the vinegar in terms of percentage of IIC 2 H 3 O2. 
Explain the chemistry involved to give the U-shaped appearance of the curve 
prior to reaching the equivalence point. What would be the general appearance 
of the graph if 0.500 N NaOH had been substituted for the NH 4 OH in the titra- 
tion? What is the advantage of using NH 4 OH? 

787. A solution approximately 0.01 N in sodium acetate is titrated con- 
ductometrically with 1.25 N HOI and the following relative conductivities 
were obtained at the corresponding buret readings: 0.00 ml. 451; 1.00 ml. 
455; 2.00 ml. - 459; 2.50 ml. - 460; 2.75 ml. - 462; 3.00 ml. - 465; 3.25 
ml. - 472; 3.50 ml. - 482; 3.75 ml. - 497; 4.00 ml. - 515; 4.50 ml. - 575; 
5.00 ml. 643; 6.00 ml. 776. Plot the curve and calculate the number 
of grams of NaC 2 H 3 O 2 present in the solution. 

788. A solution containing sodium bromide is titrated with 0.650 N silver 
acetate. The following relative values were obtained for the conductances of 
the solution during the titration. Plot the curve on a large scale and calculate 
the number of grams of NaBr originally present in the solution. 0.00 ml. 
269; 0.50 ml. - 262; 1.00 ml. - 241; 1.50 ml. - 227; 2.00 ml. - 213; 2.50 
ml. - 197; 3.00 ml. - 218; 3.50 ml. - 237; 4.00 ml. - 261; 4.50 ml. - 282; 
5.00 ml. - 301. 

789. Oxalic acid is a dibasic acid and can be considered as an equimolar 
mixture of a fairly strong acid (H 2 C 2 O 4 + 11+ + IlC 2 Or; K' = 4 X 10" 2 ) 
and a weak acid (HC 2 O 4 ~ * H+ + C 2 O 4 -; K" = 5 X 10~ 6 ). The following 
relative conductance values were obtained in the conductometric titration of a 
dilute solution of oxalic acid with 0.640 N NH 4 OH. Plot the curve and cal- 
culate the number of grams of H 2 C 2 O 4 .2H 2 O present in the solution. 0.00 
ml. - 285; 0.20 ml. - 235; 0.40 ml. - 188; 0.60 ml. - 141; 0.70 ml. - 118; 
0.80 ml. - 109; 0.90 ml. - 115; 1.00 ml. - 123; 1.20 ml. - 147; 1.40 ml. - 
173; 1.60 ml. - 184; 1.80 ml. - 183; 2.00 ml. - 181; 2.20 ml. - 1.81. What 
does the lowest point of the curve represent? Show how it would be possible 
in certain cases to analyze a mixture of oxalic acid and sodium binoxalate 
(NaHC 2 O 4 ). What would be the general appearance of the curve in this case? 



CHAPTER XVIII 
AMPEROMETRIC TITRATIONS 

108. Principle of an Amperometric Titration. Suppose a solu- 
tion containing a reducible substance is subjected to electrolytic 
reduction at an electrode consisting of metallic mercury dropping 
at a steady rate from a capillary tube, thus exposing a constantly 
fresh surface of the metal to the solution. If the applied e.m.f. 



Diffusion current 



& 

Half wave 



Residual current^ 



Applied voltage 
FIG. 12. Typical current- voltage curve (dropping-mercury electrode). 

is increased gradually, the amperage of the current remains near 
zero and increases only slightly until the decomposition potential 
of the substance is reached (see Sec. 58). At this point electrolytic 
reduction starts, and an increased e.m.f. causes a sharp increase 
in amperage in accordance with Ohm's Law, E = IR. As elec- 
trolysis progresses, however, there is a depletion of the reducible 
substance at the electrode and the potential necessary for de- 
composition is increased. A point is reached where an increasing 

275 



276 CALCULATIONS OF ANALYTICAL CHEMISTRY 

e.m,f. again causes practically no increase in amperage. These 
three steps are shown in Fig. 12 where milliamperes as ordinates 
are plotted against applied voltages as abscissas. 

The nearly constant current corresponding to the upper right- 
hand part of the curve is called the diffusion current, and its 
magnitude is proportional to the concentration of the reducible 
substance remaining in the solution. 




Volume of titrating reagent 
FIG. 13. Amperometric titration curve (Pb+~*~ + SO 4 ~). 

Suppose a reducible substance in a solution is subjected to an 
initial e.m.f. which is of such magnitude as to give the amperage 
corresponding to the diffusion current, and suppose the solution is 
subjected to a precipitation titration with a nonreducible reagent. 
Such a case would be the titration of a solution of a lead salt with 
sulfate. The concentration of the reducible substance (Pb" 1 " 1 ") is 
steadily diminished as the sulfate is added. The current, being 
proportional to the concentration of lead ions, likewise steadily 
diminishes as the titration proceeds and, if amperes are plotted 
(as ordinates) against volume of titrating solution added (as ab- 
scissas), a curve similar to that of Fig. 13 is obtained. The 
equivalence point corresponds to the point of intersection of the 



AMPEROMETRIC TITRATIONS 



277 



two arms of the curve. These arms are essentially straight lines, 
especially if the effect of dilution is corrected for. In the above 
case, the nearly horizontal portion of the curve corresponds to the 
diffusion current of a saturated solution of lead sulfate. Solu- 
bility effects may give a curved line in the close neighborhood of 
the equivalence point, but, as in certain conductometric titration 
curves [see Fig. 10(c)3, extension of the two straight parts of 




-Dropping 
mercury electrode 

Hg 



* 

NAAAAA/WVWW 



Electrolysis 
cell 



FIG. 14. Amperometric titration hook-up. 

the curve will give a point of intersection corresponding to 
the equivalence point. As in conductometric titrations, it is ad- 
vantageous to titrate a dilute solution with a relatively con- 
centrated one. 

An equally satisfactory titration curve is given in the ampero- 
metric titration of a nonreducible ion when it is titrated with a 
reagent capable of electrolytic reduction. This type is illustrated in 
the first part of the problem given below. The second part of the 
problem illustrates a titration in which both the substance titrated 
and the reagent yield a diffusion current at the applied e.m.f. 

Certain titrations involving neutralization, oxidation, and re- 
duction, and complex-ion formation are also possible by ampero- 
metric methods. 



278 CALCULATIONS OF ANALYTICAL CHEMISTRY 

Current-voltage curves can be automatically recorded on photo- 
graphic paper by means of a mechanism called a polarograph, but 
satisfactory curves can also be obtained manually with a relatively 
simple dropping-electrode assembly which furnishes a means of 
applying a variable known e.m.f. to the cell and for measuring 
the resulting very small current. Several portable instruments of 
this type are on the market. The essential hook-up is shown 
diagrammatically in Fig. 14. 

In general, although much work is still needed to bring about 
refinements of method and to extend the applications of the proc- 
ess, amperometric titrations are in many specific cases capable of 
giving very precise results. The method is satisfactory in many 
precipitation titrations where the solubility of the precipitate is 
too great for potentiometric or indicator methods to be used. 
Furthermore, foreign electrolytes which are often harmful in con- 
ductometric titrations do not usually interfere in amperometric 
titrations unless they are present at high concentrations and yield 
diffusion currents at the applied e.m.f. 

The numerical values in the following problem are taken from 
data and graphs obtained by I. M. Kolthoff and his co workers. 

Problem 

790. In each of the following two titrations, plot the titration graph and 
determine the volume of titrating solution corresponding to the equivalence 
point. Calculate the number of grams of titrated constituent shown to be 
present in each case and explain the appearance of the curve. 

(a) A certain volume of 0.0100 M K 2 SO 4 is titrated amperometrically with 
0.100 M Pb(NO 3 ) at e.m.f. = -1.2 volts. The following values are the 
milliamperes obtained at the corresponding volumes of titrating solution. 
Dilution effects have been corrected for. 0.0 ml. = 0.8; 1.0 ml. = 0.8; 2.0 ml. = 
0.8; 3.0 ml. = 0.8; 4.0 ml. = 0.9; 4.5 ml. = 1.3; 5.0 ml. = 4.2; 5.5 ml. = 
11.3; 6.0 ml. = 20.0; 6.5 ml. - 28.9; 7.0 ml. = 37.5. 

(6) A 50-ml. pipetful of dilute Pb(NO 3 ) 2 (in 0.10 M KNO 3 ) is titrated with 
0.0500 M K 2 Cr 2 O 7 at e.m.f. = - 1.0 volt. The following values are the milli- 
amperes actually obtained at the corresponding volumes of titrating solution. 
Before plotting the titration curve, correct for dilution effect by multiplying 
each current reading by (V + v}/V in which V = initial volume of solution, 
and v = total volume of reagent added. 0.0 ml. = 81.56; 1.0 ml. = 66.22; 
2.0 ml. = 48.34; 3.0 ml. = 31.66; 4.0 ml. = 15.25; 4.8 ml. = 3.79; 4.9 ml. = 
2.09; 5.0 ml. = 2.9; 5.1 ml. = 5.1; 5.3 ml. = 12.03; 5.5 ml. = 21.86; 6.0 ml. = 
43.86. (Note. The large residual current at the equivalence point is due to 
the relatively high solubility of the precipitated PbCrO 4 in the acid formed 
by the titration: 2Pb++ + Cr 2 7 " + H 2 O -> 2PbCrO 4 + 2H+.) 



PART V 

GAS ANALYSIS 

CHAPTER XIX 
CALCULATIONS OF GAS ANALYSIS 

109. Fundamental Laws. Problems involving the determina- 
tion of the proportional amounts of the components of a gaseous 
mixture and the determination of the amount of a given sub- 
stance by measuring the quantity of gas which that substance 
may be made to evolve in chemical reaction are the only phases 
of gas analysis considered in this book. 

Calculations of gas analyses make use of the following gas laws, 
most of which apply strictly only to the so-called "perfect" gases, 
but which may be applied to ordinary analyses with results that 
are usually in keeping with the precision of analytical manipulation. 
These laws should already be more or% less familiar to the student. 

Boyle's Law. The volume of a fixed mass of a gas at constant 
temperature is inversely proportional to the pressure to which it 
is subjected. That is, 

pv = p'v' = k 

where pv and p'v' are pairs of simultaneous values of pressure 
and volume of a given mass of gas and k is a constant. 

EXAMPLE I. If a sample of gas occupies a volume of 500 ml. 
at a barometric pressure of 755 mm. of mercury, what volume 
would it occupy at a pressure of 760 mm.? 

SOLUTION: An increase in pressure must cause a decrease in 
volume. In this case, the new volume will be 



500 X = 49G.7 ml. Ans. 
7oU 

Or, by substitution in the formula above, 

755 X 500 = 760 X x 
whence, 

x = 496.7 ml. Ans. 
279 



280 CALCULATIONS OF ANALYTICAL CHEMISTRY 

Charles's Law. The volume of a fixed mass of a gas at con- 
stant pressure is directly proportional to the absolute temperature 
to which it is subjected; that is, 

L = L 

v 1 " T' 

where vT and v'T' are pairs of simultaneous values of volume 
and temperature expressed on the absolute scale. Zero on the 
absolute scale is at 273C.; hence, the temperature in absolute 
units may be found by adding 273 to the temperature in centi- 
grade units. Charles's law may therefore be written 

v = 273 + t 
v' ~ 273 + // 

where t and t f represent the respective temperatures in degrees 
centigrade. 

EXAMPLE II. If a gas occupies a volume of 500 ml. at 20C. 
and the temperature is raised to 30C. at constant pressure, what 
is the new volume of the gas? 

SOLUTION : The temperatures on the absolute scale are 293 and 
303C., respectively. If the temperature is raised, the gas must 
expand and the new volume becomes 

OfkO 

500 X ~ = 517 ml. Am. 



Or, by substitution, 

500 = 273 + 20 

x "273 + 30 
whence, 

x = 517 ml. Ans. 



. . 

The formulas expressing the two gas laws mentioned above may 
combined to ive 



be combined to give 



pv __ 
"T "" 



Dalton's Law. The pressure exerted by a mixture of gases is 
equal to the sum of the pressures of the individual components, 
and the pressure exerted by a single component is the same as 
the pressure that component would exert if existing alone in the 
same volume. 



CALCULATIONS OF GAS ANALYSIS 281 

EXAMPLE III. Moist hydrogen gas is confined over water under 
a pressure of 760 mm. of mercury and a temperature of 26C. 
What is the actual pressure of the hydrogen? 

SOLUTION: At 26C., the vapor pressure of water is equal to 
25 mm. of mercury (see Table V, Appendix). The partial pressure 
of the hydrogen is therefore 760 25 = 735 mm. Ans. 

Gay-Lussac's Law. Whenever gases unite or gaseous products 
are formed, the proportions by volume measured at the same 
temperature and pressure of all the gaseous products concerned can 
be represented by ratios of small integers. Thus, in the reaction 

2H 2 + O 2 -> 2H 2 O 

two parts by volume of hydrogen unite with one part by volume 
of oxygen to give two parts by volume of water vapor. 

Avogadro's Law. Equal volumes of all gases under identical 
conditions of temperature and pressure contain the same number 
of molecules. 

110. Gas-volumetric Methods. For convenience, gas analysis 
may be divided into the following groups: 

(a) Gas-volumetric methods 

(b) Absorption methods 
(r) Combustion methods 

Under gas-volumetric methods may be included those methods 
in which a gas is evolved by means of a chemical reaction, and 
from the volume of the gas the weight of the substance producing 
it is calculated. 

From Avogadro's law it is evident that the weights of equal 
volumes of gases will be in direct proportion to the respective 
molecular weights. The weight in grams of 22.4 liters of any 
gas, when measured under standard conditions, i.e. 7 at 0C. and 
under a pressure of 760 mm. of mercury, represents the molecular 
weight of the gas. If the molecular weight of a gas and the vol- 
ume that a certain quantity of it occupies under standard conditions 
are known, the weight of that quantity can be readily determined. 
This is the principle underlying gas-volumetric analysis. Since it 
is usually inconvenient actually to measure the volume of a gas 
at 0C. and under 760 mm. pressure, it is customary to measure 
the gas at any convenient temperature and pressure and by means 



282 CALCULATIONS OF ANALYTICAL CHEMISTRY 

of Boyle's and Charles's laws to calculate the volume that the gas 
would occupy under standard conditions. 

EXAMPLE. A gas occupies a volume of 42.06 ml. under 765.0 
mm. pressure and at 20.0C. What is its volume under standard 
conditions? 

SOLUTION : According to Boyle's law, if the pressure of the gas 
at a constant temperature is reduced from 765 to 760 mm., the 
volume would be increased in the same ratio, and, were the tem- 
perature the same, the new volume would be 



42.06 X = 42.34 ml. 



The temperature, however, is to be reduced from 20.0C. (293 
Absolute) to 0C. (273 Absolute), and, according to Charles's 
law, this change alone serves to decrease the volume of the gas 
by the ratio of 293 : 273. If both the pressure and temperature are 
changed to standard conditions, the volume of the gas becomes 



Expressed according to the symbols used above, 



v x P. x 
VX ' X 



p' 273+ ~ 
which is identical to the general expression 



111. Correction for Water Vapor. Evolved gases are frequently 
collected and measured over liquids which exert an appreciable 
vapor pressure, and in such cases the barometric pressure does 
not represent the pressure of the pure gas. It may be assumed 
that the gas will be saturated with the vapor of the liquid over 
which it is measured, and in such cases the vapor pressure of the 
liquid depends only upon the temperature. According to Dalton's 
law, the pressure of the pure gas may be found simply by sub- 
tracting the vapor pressure of the liquid at the given temperature 
from the barometric pressure. The values of the vapor pressure 
of water at different temperatures are given in Table V (Appendix). 



CALCULATIONS OF GAS ANALYSIS 283 

112. Calculations of Gas-volumetric Analyses. These consider- 
ations may be applied to determine the percentage of a constituent 
of a given substance by gas-volumetric measurements. 

EXAMPLE. A 0.500-gram sample of limestone on treatment 
with acid liberates 98.7 ml. of carbon dioxide when measured over 
water at 23C. and 761 mm. pressure. What is the percentage 
of CO 2 in the sample? 
SOLUTION : 

Vapor pressure of water at 23C. = 20.9 mm. 
Pressure of the pure CO 2 = 761 - 20.9 = 740 mm. 
Volume of CO2 under standard conditions = 

740 273 



The gram-molecular weight (44 grams) of CO 2 would occupy under 
standard conditions a volume of 22.4 liters = 22,400 ml. The 
weight of C02 evolved is, therefore, 



gram 



Percentage of CO 2 in the sample is 
0.174 



0.500 



X 100 = 34.8 per cent. Ans. 



Alternative Method. Some chemists prefer to solve problems 
involving molar relationships of gases by means of the following 
general formula: 

pv = NRT. 

where p = pressure of the gas, atmospheres 

_ pressure in millimeters 



v = volume, milliliters 
N = number of moles of gas 
__ weight of gas 

molecular weight 
R = "gas constant " = 82.07 
T = temperature on the absolute scale 



284 CALCULATIONS OF ANALYTICAL CHEMISTRY 
Applying this formula to the problem under consideration, 

740 v OQ 7 _ wt - of g as v co n7 N/ OOP 
760 X y8 ' 7 ~ 44.0 X ^' U ' X ^ b 

Weight of C0 2 = 0.174 gram 

Percentage = ' .^ X 100 = 34.8 per cent. Ans. 

U.OUU 

Problems 

791. If 500 ml. of hydrogen gas are cooled at constant pressure from 26 
to 10C., what is the volume at the lower temperature? 

440 ml 



792. The pressure on a gas that at 758 mm. occupies a volume of 600 ml. 
is increased to 774 nun. at constant temperature. What is the resulting 
volume? 

Ans. 588 ml. 

793. Three hundred and sixty volumes of hydrogen are measured dry at 
13CX and 760 mm. pressure. By heating at constant pressure, the volume 
is increased 10 per cent. What is the increase in temperature? 

Ans. 26C. 

794. One hundred grams of pure calcium carbonate are dissolved in hydro- 
chloric acid. Calculate the volume of gas evolved (a) measured dry at 0C. 
and 760 mm. pressure, (6) measured dry at 15C. and 780 mm. pressure, 
(c) measured over water at 30C. and 748 mm. barometric pressure. 

Ans. (a) 22 A liters, (b) 23.0 liters, (c) 26.4 liters. 

796. How many liters of oxygen gas measured over water at 17C. and 777 
mm. pressure can be obtained from 1.00 kilogram of pure KC1O 3 by ignition 
to KC1? 

Ans. 290 liters. 

796* What weight of CaCOa must be treated with acid to produce 138.6 
ml. of CO 2 , measured over water (saturated with CC>2) at 10C. and 773 mm. 
pressure? 

Ans. 0.599 gram. 

797. In the analysis of dolomite, 0.0500 gram of ferric oxide, 0.6080 gram 
of CaO, and 0.1505 gram of magnesium pyrophosphate were obtained. If 
these were originally present as FeCOs, CaCO 3 , and MgCOa, how many milli- 
liters of CO2 measured dry at 20C. and 780 mm. pressure could have been 
obtained from the same weight of sample? 

Ans. 300.3ml. 



CALCULATIONS OF GAS ANALYSIS 285 

798. What weight of limestone should be taken for analysis such that the 
volume in milliliters of CO 2 measured dry at 20C. and 780 mm. equals the 
percentage of CO 2 present? 

Ans. 0.1880 gram. 

799. A sample of pyrite (FeS 2 ) weighing 0.2000 gram yields 0.7783 gram 
of BaSO 4 . How many cubic feet of air measured at 130F. and 27 in. of mer- 
cury pressure would theoretically be required to burn 1.00 pound of the 
pyrite? What would be the volume of the gaseous residue (sulfur dioxide 
and residual nitrogen) measured at the same temperature arid pressure? 
(4FeS 2 + 11O 2 -> 2Fe 2 3 + 8SO 2 . Air = 20.9 per cent O 2 by volume. 1 cu. 
in. = 16.39 ml. 1 Ib. = 0.4536 kg.) 

Ans. 52.4 cubic feet. 49.5 cubic feet. 

800. Compute the volume of H 2 O that can be obtained from 8.0 grams of 
I^Ca^AUSiioO^ measured at (a) 20C. and 750 mm. pressure, (6) 750 mm. 
pressure and 900C. (Two significant figures.) 

Ans. (a) 0.18 ml., (fc) 970 ml. 



801. If a gas measured dry at 27C. and 758 mm. pressure occupies a 
volume of 500 ml., calculate its volume if the temperature is increased to 87C. 
and the pressure is kept constant. 

802. If hydrogen gas when measured over water at 23C. and 772 mm. 
pressure occupies 97.3 ml., what would b the volume under standard 
conditions? 

803. A gas occupies a volume of 222 ml. over water at 12C. and 751 mm. 
pressure. What volume would it occupy over water at 31 C. and 770 mm. 
pressure? 

804. BaCO 3 and MgCO 3 are mixed in the proportions by weights of 2:1. 
Calculate the volume of 6.00 N HC1 to decompose a 5.00-gram sample. Cal- 
culate the volume of CO 2 gas formed when measured dry at 22.4C. and 
758 mm. pressure. What would the volume of the gas be if it were collected 
under the same conditions over water (saturated with CO 2 )? 

805. What weight of impure calcite (CaCOs) should be taken for analysis 
so that the volume in milliliters of CO 2 obtained by treating the sample with 
acid and measuring the CO 2 dry at 18C. and 763 mm. pressure will equul the 
percentage of CaO in the sample? 

806. What volumes of nitrogen and carbon dioxide, each measured dry 
at 20C. and 755 mm. pressure, could be obtained by the combustion of 
0.2010 gram of urea [CO(NH 2 ) 2 ]? 

807. What volume of nitrogen measured over water at 30C. and 760 mm. 
pressure could be obtained from 0.1860 gram of tetraethyltetrazone 

l(C 2 H 6 ) 2 :N.N:N.N:(C 2 H 6 ) 2 ]? 



286 CALCULATIONS OF ANALYTICAL CHEMISTRY 

808. If, in the analysis of a 1.00-gram sample of a carbonate, 18.0 ml. of 
CO 2 measured over water at 18C. and 763 mm. pressure were obtained, find 
the percentage of carbon in the sample. 

809. What weight of limestone should be taken for analysis so that the 
volume of CC>2 evolved measured over water at 15C. and 749 mm. pressure 
shall be three-fifths the percentage of CC>2 in the sample? 

810. Compute the volume of oxygen required to oxidize a sample of pure 
Fe weighing 0.9000 gram, assuming that the product of combustion is com- 
posed of 60 per cent Fe2O 3 and 40 per cent FeaCU and that the gas is measured 
dry at 21C. and 756 mm. pressure. 

811. A compound of C, N, and II yields a volume of nitrogen which when 
measured in milliliters over water at 22C. and 767 mm. pressure is equal to 
155.5 times the number of grams of sample taken. The carbon and hydrogen 
are present in the molar ratio of 1:1. What is the empirical formula of the 
compound? 

812. Decomposition of a compound of carbon, hydrogen, nitrogen, oxygen, 
and bromine weighing 0.2000 gram yielded 8.70 ml. of nitrogen, measured 
over water at 18C. and 758 mm. pressure. Combustion in oxygen of the 
same weight of sample gave 0.1880 gram CO 2 and 0.01924 gram H 2 O. After 
decomposition of 0.2000 gram with HNO 3 , a precipitate of AgBr weighing 
0.2674 gram was obtained. The molecular weight was found to be about 275. 
What is the formula of the compound? 

813. Decomposition of 0.1500 gram of indole gave 16.42 ml. of nitrogen 
when measured over water at 27C. and 758 mm. pressure. Combustion in 
oxygen of 0.2000 gram of the sample increased the weight of a potash bulb by 
0.6026 gram and of a calcium chloride tube by 0.1078 gram. Calculate the 
empirical formula of indole. 

113. Absorption Methods. Absorption methods of gas analy- 
sis apply to the determination of the proportionate amounts of 
the components of a gaseous mixture. The mixture of gases is 
treated with a series of absorbents, and the temperature and pres- 
sure are usually kept constant throughout the entire determina- 
tion. In cases where these are allowed to vary, corrections for 
their effect may be made by applying the principles outlined in 
Sec. 109. The difference in the volume of the gas before and 
after it has been acted upon by each absorbing agent represents 
the amount of gas absorbed, and the amount is usually expressed 
on a percentage-by-volume basis. The many forms of apparatus 
used for carrying out gas absorptions are described in the text- 
books on the subject, but the fundamental principles are identical. 
The reagents commonly employed are shown below. 



CALCULATIONS OF GAS ANALYSIS 287 

GAS REAGENT 

Carbon dioxide Caustic soda 

Caustic potash 

Unsaturated hydrocarbons Bromine water 

(" illuminants") Fuming sulfuric acid 

Oxygen Alkaline pyrogallol solution 

Yellow phosphorus 

Carbon monoxide Ammoniacal cuprous chloride 

Hydrogen Palladium sponge 

Palladous chloride solution 
Colloidal palladium solution. 

EXAMPLE. A sample of illuminating gas occupying a volume 
of 80.0 ml. is treated in succession with caustic potash solution, 
fuming sulfuric acid, alkaline pyrogallol solution, and ammo- 
niacal cuprous chloride solution. After each treatment, the vol- 
ume of the residual gas at constant temperature and pressure is 
measured as 78.7, 75.5, 75.1, and 68.3 ml., respectively. What 
is the percentage composition of the gas as shown by these results? 
SOLUTION: 

Volume of CO 2 = 80.0 - 78.7 = 1.3 ml. 

Volume of illuminants = 78.7 75.5 = 3.2 ml. 

Volume of 2 = 75.5 ^ 75.1 = 0.4 ml. 

Volume of CO = 75.1 - 68.3 - 6.8 ml. 

The percentages of the various components are therefore 
^ X 100 =1.6 per cent CO 2 

oU.U 

3 2 
5777: X 100 = 4.0 per cent illuminants 

oU.U 



0.4 
80.0 



X 100 = 0.5 per cent 2 



r X 100 = 8.5 per cent GO 

oU.U 
f\R o 



577 

oU.U 



X 100 = 85.4 per cent inert gases 



Ans. 



114. Combustion Methods. If a gas mixture contains one or 
more components capable of combustion with oxygen, it is usually 
possible to determine the percentages of these components by 
allowing combustion to take place and measuring the contraction 



288 CALCULATIONS OF ANALYTICAL CHEMISTRY 

in volume, the amount of carbon dioxide formed, the volume of 
oxygen used, or combinations of these measurements, depending 
upon the number and character of the combustible components 
present. Gay-Lussac's law underlies calculations involving con- 
tractions in volume. Thus, in the combustion of carbon monoxide 
with oxygen 

2CO + 2 -> 2C0 2 

two volumes of carbon monoxide unite with one volume of oxygen 
to form two volumes of carbon dioxide. The combustion is there- 
fore accompanied by a contraction equal to one-half the volume 
of the carbon monoxide present and produces a volume of carbon 
dioxide equal to the original volume of carbon monoxide. 

Assume a gas mixture with hydrogen and methane as the only 
combustible components. Hydrogen reacts with oxygen accord- 
ing to the equation 



in which two volumes of hydrogen unite with one volume of oxygen 
to form water vapor, condensing at ordinary temperatures to 
liquid water. Methane reacts with oxygen according to the equa- 
tion 

CH 4 + 20 2 -> CO 2 + 2H 2 

in which one volume of methane reacts with two volumes of oxygen 
to form one volume of carbon dioxide. Let x represent the volume 
of hydrogen and y the volume of methane present in the gas 
mixture. The volume of oxygen required for the hydrogen is ^z, 
and the volume of oxygen required for the methane is 2y. The 
total volume of oxygen required B is therefore given by the ex- 
pression 
0) B=y 2 x + 2y 

The contraction in volume caused by the hydrogen reaction is 
3 x, and that by the methane reaction is 2y. The total contrac- 
tion in volume C is given by the expression 

(2) C = 3^ + 2y 

whence 

x = C B = volume of hydrogen 



1 r .u 

y = : = volume of methane 



CALCULATIONS OF GAS ANALYSIS 289 

It is evident that by allowing this gas mixture to react with a 
determinable volume of oxygen and measuring the resulting con- 
traction which the gas undergoes it is possible to determine the 
volume of hydrogen and methane present. 

Since carbon dioxide is appreciably soluble in water, it is cus- 
tomary in accurate analyses to measure the contraction in volume 
after the carbon dioxide has been entirely absorbed. Under such 
conditions, in the combustion of a mixture of hydrogen and 
methane the volume of oxygen required would be represented as 
before by the equation 

(1) B = y 2 x + 2y 

but the total decrease in volume due to combustion and absorption 
would be 

(2) C" = %x + 3y 
whence 

x = %C' 2B = volume of hydrogen 
y = B ]/^C f = volume of methane 



Instead of measuring the contraction in volume and the oxygen 
consumed, the amounts of hydrogen and methane present in a 
mixture in which they are the only combustible components may 
be determined from the contraction in volume and the volume of 
carbon dioxide produced by combustion. Combustion of hydro- 
gen of volume x causes a contraction in volume of % x and pro- 
duces no carbon dioxide; combustion of methane of volume y 
causes a contraction of 2y and produces a volume of carbon dioxide 
equal to y. The total contraction in volume C is therefore given 
by the equation 

C = %x + 2y 



and the total volume of carbon dioxide produced D is given by 

D = y 
Hence, 

2C-4/) , ,,, 

x = - ^ - = volume of hydrogen 

o 

y = D = volume of methane 

In a similar way, the percentage composition of other mixtures 
of gases may usually be calculated, provided that as many inde- 



290 



CALCULATIONS OF ANALYTICAL CHEMISTRY 



pendent equations can be formulated as there are unknown com- 
ponents in the mixture. 

The equations in the following table represent combustion re- 
actions more commonly encountered in gas analysis, and the ac- 
companying columns show the volume relationships in each 
case. 





Vol. 
gas 


02 

con- 
sumed 


Con- 
trac- 
tion 


CO. 

pro- 
duced 


Hydrogen 2Ha + Oa 2IIzO 


1 


H 


* 1H 





Carbon monoxide 2CO + Oa > 2CO2 


1 


yz 


H 


1 


Methane CHi + 2Oa -> COa -f 2H 2 O 


1 


2 


2 


1 


Acetylene 2C2H a 4- 5O2 - 4CO2 -f 2H 2 O 


1 


2K 


1H 


2 


Ethylene C2lli + 3O2-* 2CO 2 4- 2H 2 O 


1 


3 


2 


2 


Ethane 2CsHo + 7O2 > 4COa 4- 6H2O 


1 


3^ 


2H 


2 


Propylene 2CsH8 -f- 9O2 6CO2 4- 6TT2O 


1 


4^ 


2H 


3 


Propane CsHs 4- 6O2 > 3CO + 4H2O 


1 


5 


3 


3 


Butane 2C4Hio + 13O* 8COa 4- 10H2O 


1 


6H 


3H 


4 













With this table, little difficulty should be experienced in formu- 
lating the necessary equations for the determination by combus- 
tion of any mixture of the gases. 

In case air is used for combustion, it may be assumed to consist 
*of 20.9 per cent of oxygen by volume. 

EXAMPLE I. A mixture of carbon monoxide and nitrogen occu- 
pies a volume of 100 ml. and on combustion with oxygen produces 
40 ml. of carbon dioxide. Calculate the percentage of nitrogen in 
the mixture. 

SOLUTION: Let x represent the volume of carbon monoxide 
and ?/ the volume of nitrogen. Since I volume of carbon monoxide 
on combustion gives 1 volume of carbon dioxide, the volume of 
carbon dioxide produced is equal to x. This volume is stated to 
be 40 ml., and the volume of nitrogen is therefore 60 ml. N 2 = 
60 per cent. Arts. 

EXAMPLE II. A mixture of carbon monoxide, methane, and 
nitrogen occupies a volume of 20 ml. On combustion with an 
excess of oxygen, a contraction of 21 ml. takes place, and 18 ml. 
of carbon dioxide are formed. What is the volume of each com- 
ponent in the mixture? 



CALCULATIONS OF GAS ANALYSIS 291 

SOLUTION: 

Let x = volume of CO 
y = volume of CTT 4 
c = volume of N% 

Total contraction in volume, C = ]/^x + 2y 
Total volume of CO 2 produced, D = x + y 

= 4D ~ 2C = 72 ~ 42 
x- 3 3 ion 

2C - D 42-18 Q , 
,,.__,__- 8 ml. 

z = volume original gas (x + y) = 2 ml. , 

EXAMPLE III. The residual gas mentioned in the example in 
Sec. 113 is assumed to consist entirely of hydrogen, methane, and 
nitrogen. To a 20.0-ml. sample are added exactly 100.0 ml. of 
air, and the mixture is exploded. After the carbon dioxide is 
absorbed in caustic potash, the volume of the gas is found to be 
88.0 ml.; and after the excess oxygen is absorbed in pyrogallol, 
the volume of the gas is 82.1 ml. What is the percentage of each 
component in the gas mixture and in the original illuminating gas? 
SOLUTION: 

Volume after adding air = 120 ml. 

Contraction after explosion and absorption = 120 88.0 = 
32.0 ml. = C" 

Volume of oxygen taken = 100.0 X 0.209 = 20.9 ml. 

Volume of residual oxygen = 88.0 82.1 = 5.9 ml. 

Oxygen actually required = 20.9 5.9 = 15.0 ml. = B 
On substitution of these values of B and C' in the equations de- 
rived above, viz., 

x = ^C' - 25 
y = B - 



the results obtained are 

x = (% X 32.0) - (2 X 15.0) = 12.7 ml. = volume of H 2 
y = 15.0 - (M X 32.0) = 4.3 ml. = volume of CEU 
20.0 - (12.7 + 4.3) = 3.0 ml. = volume of N 2 

The percentages by volume of these components are found by 
dividing these volumes by 20.0 and multiplying by 100. 



292 CALCULATIONS OF ANALYTICAL CHEMISTRY 

63.5 per cent H2 1 

21.5 per cent CH4 f Ans. 

15.0 per cent N 2 J 

In the original illuminating gas (Example, Sec. 113) the percent- 
ages of these components are 



Ans. 



12.7 X ^~ X ~ = 54.2 per cent H 2 

AS Q inn 
4.3 X ~ X ^ = 18.3 per cent CH 4 



3.0 X ~ X = 12.8 per cent N 2 



Problems 

814. The following measurements are made under identical conditions. 
Calculate the percentages of CO 2 , O 2 , CO, and N 2 in a sample of gas contain- 
ing no other components. 

Sample taken = 100.0 ml. 

Volume after KOII treatment = 91.5 ml. 

Volume after pyrogallol treatment = 81 .4 ml. 

Volume after cuprous chloride treatment = 81.1 ml. 

Ans. CC>2 = 8.5 per cent, O 2 = 10.1 per cent, CO = 0.3 per cent, N 2 = 
81.1 per cent. 

815. A flue gas is known to contain 3.8 per cent O 2 , 0.6 per cent CO, 
15.0 per cent CO 2 , and the rest N 2 . A 95.0-ml. sample is drawn into an 
Orsat apparatus. What would be the volume reading after absorption in the 
following absorbents in the order stated: (a) caustic potash, (6) pyrogallol, 
(c) ammoniacal cuprous chloride? 

Ans. (a) 80.7 ml., (6) 77.1 ml., (c) 76.5 ml. 

816. How many liters of oxygen are necessary for the complete combustion 
of 5.0 liters of (a) methane, (6) acetylene, (c) hydrogen sulfide? 

Ans. (a) 10 liters, (6) 12.5 liters, (c) 7.5 liters. 

817. The following measurements are made under identical conditions. 
Calculate the percentage composition of a mixture of hydrogen and nitrogen. 

Volume of gas taken = 58.2 ml. 

Volume of oxygen added = 32.0 ml. 

Volume of oxygen consumed by combustion = 6.1 ml. 

Ans. H 2 = 21.0 per cent, N 2 = 79.0 per cent. 

818. The following measurements are taken under identical conditions. 
Calculate the percentage composition of a mixture of hydrogen and nitrogen. 



CALCULATIONS OF GAS ANALYSIS 293 

Volume of gas taken = 95.3 ml. 

Volume of oxygen added = 40.8 ml. 

Volume of gas after combustion = 40.1 ml. 

Ans. H 2 = 67.2 per cent, N 2 = 32.8 per cent. 

819. What is the percentage composition of a mixture of hydrogen and 
nitrogen if the contraction in volume due to combustion with oxygen is the 
same as the volume of the sample taken? 

Ans. II 2 = 66?^ per cent, N 2 = 33> per cent. 

820. What is the percentage of methane in a mixture of hydrogen, methane, 
and acetylene if 16.0 ml. of the mixture when exploded with an excess of air 
cause a contraction of 26.0 ml.? 

Ans. 25.0 per cent. 

821. The following measurements are made under identical conditions. 
Calculate the percentage composition of a mixture of hydrogen, carbon 
monoxide, and methane. 

Volume of gas taken = 10.5 ml. 

Volume of air added = 137.4 ml. 

Total volume after combustion = 136.1 ml. 
Volume after removing CO 2 = 129.6 ml. 

Ans. H 2 = 38.1 per cent, CO = 45.7 per cent, CII 4 = 16.2 per cent. 

822. What is the percentage of propane in^ a mixture of propane, carbon 
monoxide, and methane if a 13.7-ml. sample on combustion produces 23.7 ml. 
of carbon dioxide? 

Ans. 36.5 per cent. 

823. What is the percentage composition of a mixture of carbon monoxide, 
ethane, and nitrogen, if, on combustion with oxygen, the contraction in 
volume and the volume of carbon dioxide produced are each numerically 
equal to the volume of the sample taken? 

Ans. CO = 33)^ per cent, C 2 H 6 = 33^ per cent, N 2 = 33^ per cent. 

824. To 40.8 ml. ,of a mixture of hydrogen, nitrogen, and carbon monoxide 
are added 150.0 ml. of air, and the mixture is exploded. If 4.8 ml. of CO 2 
are produced and the residual oxygen requires 42.0 ml. of hydrogen for 
combustion, what is the percentage composition of the original mixture, 
and what was the total volume after the first combustion? 

Ans. H 2 = 39.0 per cent, N 2 = 49.3 per cent, CO =11.7 per cent. 
Volume = 164.6 ml. 

826. A mixture of ethane (C 2 H 6 ), hydrogen, carbon monoxide, and nitrogen 
has a volume of 28.0 ml. After combustion with 72.0 ml. of oxygen, the 
residual volume is 60.0 ml. and, after passing this into KOH solution, the 
residual gas occupies 34.0 ml. When this gas is passed over yellow phosphor- 



294 CALCULATIONS OF ANALYTICAL CHEMISTRY 

ous, only 4.0 ml. are left. Calculate the percentage composition of the original 



Q. 

Ans. C 2 H 6 = 35.7 per cent, H 2 = 28.6 per cent, CO = 21.4 per cent, 
N 2 = 14.3 per cent. 

826. What is the percentage composition of a mixture of hydrogen, carbon 
monoxide, and methane if the volume of the oxygen consumed in combustion 
and the volume of the carbon dioxide produced are each equal to three- 
fourths of the volume of the original gas taken? 

Ans. 112 = 25 per cent, CO = 58.3 per cent, CH 4 ~ 16.7 per cent. 

827. From the following data, calculate the percentage composition of a 
sample of illuminating gas: 

Sample taken for analysis = 100.6 ml. 

Volume after KOH treatment = 98.4 ml. 

After Br 2 treatment = 94.2 ml. 

After pyrogallol treatment = 93.7 ml. 

After Cu 2 Cl 2 treatment = 85.2 ml. 

Residual gas taken for analysis == 10.3 ml. 

Volume of air added = 87.3 ml. 

Volume after explosion = 80.1 ml. 

Carbon dioxide produced = 5.2 ml. 

Ans. CO 2 = 2.2 per cent 
Unsaturated compounds = 4.2 per cent 

O 2 = 0.5 per cent 

CO = 8.5 per cent 

CH 4 = 42.8 per cent 

H 2 = 38.6 per cent 

N 2 = 3.3 per cent 

828. The following measurements are made under identical conditions. 
Calculate the percentages of carbon dioxide, oxygen, carbon monoxide, and 
nitrogen in a sample of gas containing no other components: 

Sample taken = 79.5 ml. 

Volume after KOH treatment = 72.9 ml. 
Volume after O 2 absorption = 64.6 ml. 
Volume after CO absorption = 64.5 ml. 

Ans. CO 2 = 8.3 per cent, O 2 = 10.4 per cent, CO = 0.1 per cent, N 2 = 
81.2 per cent. 

829. A water gas is of the following composition: 33.4 per cent CO, 8.9 
per cent unsaturated hydrocarbons, 3.9 per cent CO 2 , 7.9 per cent N 2 , 10.4 per 
cent saturated hydrocarbons, 34.6 per cent H 2 , 0.9 per cent O 2 . If a sample of 
100 ml. is passed through the following absorbents in the order stated until 
constant volume is reached in each case, what is the volume reading following 
each treatment: (a) caustic potash, (6) bromine water, (c) alkaline pyrogallol, 
(d) ammoniacal cuprous chloride? 



CALCULATIONS OF GAS ANALYSIS 295 

830. A mixture of methane, air, and hydrogen having a volume of 130 ml. is 
conducted over gently ignited palladium asbestos, after which the volume of 
the gas is 105 ml. Both measurements are made at 20C. and 750 mm. pres- 
sure, a Hempel pipet filled with water being used. Compute the percentage 
(by volume) of hydrogen in the original gas mixture and the weight of H 2 O 
that could be formed from it. 

831. If 12.0 grams of pure carbon undergo combustion in 31.3 liters of 
pure oxygen, what is the percentage-by-volume composition of the mixture 
after combustion? 

832. Assume air to contain 20.9 volumes O 2 and 79.1 volumes of nitrogen. 
If 100 volumes of air are mixed with 95 volumes of hydrogen and the mixture 
exploded, what is the composition of the gas remaining and what are the 
volumes of the various components after cooling to 20C. and 760 mm. 
pressure? 

833. A known volume of a mixture of methane, carbon monoxide, and 
nitrogen is exploded with an excess of air. Show by equations that the 
percentage composition of the mixture cannot be determined by measuring 
the contraction in volume and the volume of oxygen consumed. 

834. A certain illuminating gas is known to contain the following compo- 
nents: H 2 , CH4, CO 2 , N 2 , O 2 , CO, and unsaturated hydrocarbons. Calculate 
the percentage composition of the gas from the following data: 

Sample taken for analysis = 99.5 ml. 

Volume after KOH treatment = 97.6 ml. 
After Br 2 treatment v = 94.4 ml. 

After pyrogallol treatment = 93.8 ml. 
After Cu 2 Cl 2 treatment = 85.1 ml. 

Residual gas taken for analysis = 12.0 ml. 
Volume of O 2 added = 20.2 ml. 

Volume after explosion =11.8 ml. 

Combined CO 2 + O 2 produced = 11.4 ml. 

(i.e., absorbed in alkaline 

pyrogallol) 

836. A certain natural gas is known to contain methane, nitrogen, and 
carbon dioxide. A 50.0-ml. sample is passed into caustic potash, and the 
volume of the residual gas is found to be 49.6 ml. Of this residual gas, 20.0 ml. 
are taken and an excess of air is added. Combustion causes a shrinkage in 
the total volume of 38.4 ml. Calculate the percentage composition of the 
original gas. 

836. A blast-furnace gas is of the following composition: CO 2 = 12.5 per 
cent, CO = 26.8 per cent, H 2 = 3.6 per cent, N 2 = 57.1 per cent. If a 
100-ml. sample were passed through a solution of caustic potash, what would 
be the volume of the residual gas? If to 50.0 ml. of this residue were added 
25.0 ml. of pure oxygen and the mixture exploded, what would be the new 
volume of the gas and what would be its percentage composition? 



PART VI 
COMMON ANALYTICAL DETERMINATIONS 

The following methods are those in common use in the gravimetric and 
volumetric determinations of the more common elements and radicals. They 
are given here in barest outline principally to serve as a reference in solving 
problems in this book. Colorimetric methods and other special methods 
involving little if any stoichiometry are omitted. 

Aluminum 

Precipitated with NII 4 OII as A1(OH) 3 , ignited, and weighed as A1 2 3 . 

Precipitated with 8-hydroxyquinoline ("oxine") as A1(C 9 H6NO) 3 , ignited, 
and weighed as A1 2 03. 

Precipitated with oxine as in the preceding case arid the precipitate dis- 
solved in HC1. KBr then added and the solution titrated with standard 
KBrOa to disappearance of red color of methyl red indicator [A1(O 9 H 6 NO) 3 + 
311+ -> A1++*- + 3C 9 H 6 NOH; 5Br~ + BrOr + 611+ - 3Br 2 + 3II 2 O; 
C 9 HNOH + 2Br 2 - 2Br~ + 211+ + C 9 H 4 NBr 2 OH]. 

Ammonium 
(See under Nitrogen.) 

Antimony 

Precipitated as Sb 2 S 3 or Sb 2 S5, ignited in an inert atmosphere, and weighed 
xus 81)283. 

Precipitated as Sb 2 S 3 or Sb 2 S 6 , heated with NII 4 OH + H 2 2 , ignited in air, 
and weighed as Sb 2 4 . 

(In alloys} Left as a residue of hydrated Sb 2 06 on treating alloy with HNO 3 
<6Sb + lONOr + 1011+ -> 3Sb 2 O 5 + 10NO + 5II 2 O). Residue ignited in 
4iir to Sb 2 O 4 . Tin must be removed. 

Titrated in ice-cold HC1 solution from 3 to 5 with standard KMn0 4 
(oSbCV + 2MnOr + 12II 2 O -> 5H 3 SbO 4 + 2Mn++ + 20C1- + 91 l+). 

Titrated from 3 to 5 with standard I 2 in a solution kept nearly neutral by 
excess NaIICO 3 . The antimony is often held as a tartrate complex (SbOs* + 
I 2 + 2HCO 3 ~ -> SbCV + 21- + 2CO 2 + H,O). 

Titrated from 3 to 5 with standard KBr0 3 in IIC1 solution to disappearance 
of color of methyl orange indicator (3SbCl 4 ~ + Br<V + 9II 2 O - 3H a SbO 4 + 
Br~ + 12C1- + 9II+). 

Brought to 5-valent form and the HC1 solution treated with KI. The 
liberated I 2 titrated with standard Na 2 S 2 O 3 (H 3 SbO 4 + 21" + 4C1- + 5H+ -> 
+ I 2 + 4H 2 O). 

297 



298 CALCULATIONS OF ANALYTICAL CHEMISTRY 

Arsenic 

Precipitated as As^S 3 from 9 N HC1 with H 2 S and weighed as such. Arsenate 
precipitated from ammoniacal solution as MgNH4As0 4 .6H 2 O, ignited, and 
weighed as Mg^s^. 

Arsenate precipitated from neutral solution as Ag*AsO 4 . Precipitate dis- 
solved in HNO 3 and the Ag f titrated with standard KCNS using ferric alum 
indicator (Ag+ + CNS~ -> AgCNS). 

Titrated from 3 to 5 with standard I 2 in a solution kept nearly neutral by 
excess NaHCO 3 (AsOr + I 2 + 2HCO 3 - -* AsOr + 2I~ + 2CO 2 + H 2 O). 

Titrated in strong HC1 solution from 3 to 5 with standard KIO S (2As+++ + 
IO 3 - + Cl" + 5H 2 O - 2Il8AsO 4 + IC1 + 4H+). Free I 2 is formed as an 
intermediate product and gives violet color with chloroform. Titration is to 
disappearance of this color. 

Titrated in HC1 solution with standard KBrO 3 to disappearance of color of 
methyl orange indicator (3As + ++ + BrCV + 9II 2 O -> 3H 3 AsO 4 + Br~ + 91I+). 

(Smatt amounts) Reduced in acid solution with Zn and evolved as AsH 3 . 
The arsine oxidized to As and color compared to standards. Or AsH 3 ab- 
sorbed in measured volume of I 2 solution. Excess I 2 titrated with standard 
NaAQi (AsHs + I 2 + 4H 2 -> H 3 AsO 4 + 81" + 8H+). 

Barium 

Precipitated as BaSO 4 , ignited, and weighed as such. 

Precipitated as BaCrO 4 and weighed as such. 

Precipitated with (NH 4 ) 2 CO 3 as BaCO 3 , ignited, and weighed as such. 

Precipitated as BaCrO 4 . Precipitate dissolved in excess standard FeS0 4 
(+H 2 SO 4 ) and excess Fe f + titrated with standard KMn0 4 (BaCrO 4 + 3Fe f f + 
8H+ + S0 4 - -> 3Fe ++ + + Cr^ ++ + BaSO 4 + 4H 2 O). 

Precipitated as BaCrO 4 and precipitate dissolved in KI + dilute IIC1. 
Liberated I 2 titrated with standard Na 2 S 2 O 8 (2BaCrO 4 + 6I~ + 16II + -^ 

2Ba++ + 3I 2 + 8H 2 O). 

Beryllium 

Precipitated with NIl^OH as Be(OH) 2 , ignited, and weighed as BeO. 
Precipitated from ammoniacal solution with 8-hydroxyquinoline, as 
Be(C 9 Il6NO) 2 , ignited, and weighed as BeO. 

Bismuth 

Precipitated with H 2 S as Bi 2 S 3 and weighed as such. 

Precipitated as basic carbonate, ignited, and weighed as Bi 2 O. 

Precipitated as oxalate, (BiO) 2 C 2 O 4 , the precipitate dissolved in dilute 
H 2 SO 4 , and the oxalate titrated with standard KMnO 4 (5H 2 C 2 O 4 + 2Mn0 4 " + 
6H+ -> 10C0 2 + 2Mn ++ + 8H 2 O). 



Boron 

Borate heated with methyl alcohol and the volatile methyl borate passed 
through a weighed amount of ignited lime: 2B(OCH s )a + CaO + 3H 2 O -* 
6CH 3 OH -f Ca(BO 2 ) 2 . The material is reignited and weighed. Gain in 
weight = B 2 CV 



COMMON ANALYTICAL DETERMINATIONS 299 

Borate treated with methyl alcohol as above and the methyl borate hy- 
drolyzed: B(OCH 8 ) 3 + 3H 2 O -> H 3 BO 3 + 3CH 3 OH. The CH,OH is re- 
moved by evaporation and the H 3 BO 3 titrated with standard NaOH in the 
presence of glycerol (or other polyhydric alcohol) which forms a loose com- 
pound with the HsBOs. Only one hydrogen of H 3 BO 3 reacts. 

Bromine 

(Bromide) Precipitated as AgBr and weighed as such. 

(Bromide) (Volhard method) Precipitated as AgBr with measured amount 
of AgNOs and the excess Ag+ titrated with standard KCNS using ferric alum 
indicator (Ag+ + CNS~ -* AgCNS). 

(Bromide) Titrated with standard AgNO 3 using eosin or other adsorption 
indicator. 

(Free bromine) Excess KI added and the liberated I 2 titrated with standard 
Na 2 S 2 O 3 (Br 2 + 21" -* I, + 2Br~). 

(Bromate) Excess KI added in the presence of acid and the liberated L> 
titrated with standard Na 2 S 2 O 3 (BrO 3 ~ + 6I~ + 6H+ -> 3I 2 + Br~ + 3H 2 O). 

(Bromate) Measured amount of As 2 O 3 (dissolved in NaHCOj) added. The 
solution is acidified, boiled, neutralized with NallCOs, and the excess arsenite 
titrated with standard I 2 (BrO" + 3H 3 AsO 3 - 3HaAsO 4 + Br~; AsOs" + 
I 2 + 2HCO- - AsO 4 - + 21- + 2CO 2 + H 2 O). 

Cadmium 

Precipitated as CdS and weighed as such. 
Electrolytically deposited as Cd and weighed as such. 

Precipitated as CdS and the precipitate titrated with standard I 2 in the 
presence of IIC1 (CdS + I 2 -* Cd++ + S + 2I~). 

Calcium 

Precipitated as CaC 2 O 4 .H 2 O, ignited at low heat, and weighed as CaCO s . 

Precipitated as CaC 2 O 4 .II 2 O, ignited strongly, and weighed as CaO. 

Precipitated as CaC 2 O4.H 2 O, ignited, moistened with H 2 SO 4> reignited, and 
weighed as CaSO 4 . 

Precipitated as CaC 2 O 4 .II 2 O, the precipitate dissolved in dilute II 2 SO 4 , and 
the oxalate titrated with standard KMnO 4 (5II 2 C 2 O 4 + 2MnO 4 " + 6H+ -* 
lOCOa + 2Mn++ + 8H 2 O). 

Precipitated as CaC 2 O 4 .H 2 O with a measured amount of oxalate. The 
precipitate is filtered and the excess oxalate in the filtrate is titrated with 
standard KMnO 4 as above. 

Precipitated as CaC 2 O 4 .H 2 O, and the ignited material (CaO, or CaC0 3 , or 
CaO + CaCOs) titrated with standard acid. 

Carbon 

(In organic compounds) Substance is burned in O 2 and the C0 2 caught in 
an absorbing agent (e.g., "asoarite") and weighed. 

(In iron and steel) Alloy is burned in O 2 . The CO 2 is caught in absorbing 
agent (e.g., "ascarite," = NaOH + asbestos) and weighed. Or the COa is 
caught in a measured volume of standard Ba(OH) 2 solution and (1) the 



300 CALCULATIONS OF ANALYTICAL CHEMISTRY 

Ba(OH) 2 filtrate or supernatant liquid is titrated with standard acid or (2) the 
change in conductivity of the Ba(OH) 2 is measured. 

(C0 2 in carbonates) (Alkalimeter method) Sample is treated with acid in a 
weighed alkalimeter and the loss in weight measured. 

(C0 2 in carbonates) Sample is treated with acid and the evolved CO 2 caught 
in an absorbing agent (e.g., "ascarite") and weighed. 

(CO Z in gas mixture) CO 2 absorbed in KOH solution and the decrease in 
volume of gas mixture measured. 

(CO in gtis mixture) CO absorbed in ammoniacal cuprous chloride solution 
and the decrease in volume of gas mixture measured. Or volume change 
measured before and after combustion with O 2 . 

(Oxalate) Precipitated as CaC 2 O 4 , ignited to CaO, or CaCO 3 , or CaS0 4 
(see under Calcium), and weighed. 

(Oxalate) Titrated with standard KMnO 4 (5C 2 O 4 - + 2MnO 4 ~ + 16H+ -> 
10CO 2 + 2Mn++ + 8H 2 O). 

Cerium 

Precipitated as Ce(OH) 4 or Ce(OII) 3 or Ce 2 (C 2 O 4 ) 3 , ignited, and weighed 
as CeO 2 . 

Precipitated as Ce(IO 3 ) 4 , converted to Ce 2 (C 2 O 4 ) 3 , ignited, and weighed as 
Ce0 2 (2CeffOW4 + 24H 2 C 2 O 4 -> Ce 2 (C 2 O 4 ) 3 + 4I ? + 42CO 2 + 24II 2 O; 
2Ce 2 (C 2 O 4 ) 3 + 4O 2 - 4CeO2 + 12CO 2 ). 

Cerous oxidized to eerie with NaBiO 3 or (NH 4 ) 2 S 2 Os and excess oxidizing 
agent removed. Measured amount of FeSO 4 added and the excess ferrous 
titrated with standard KMn0 4 (2Ce+++ + NaBiO 3 + 6H+ -> 2Ce++++ + 
Na+ + 3H 2 O; 



Chlorine 

(Chloride) Precipitated as AgCl and weighed as such. 

(Chloride) (Volhard method) Precipitated as AgCl with measured amount 
of AgNO 8 and the excess Ag+ titrated with standard KCNS using ferric alum 
indicator (Ag + + CNS" -* AgCNS). 

(Chloride) Titrated with standard AgNO 3 using fluorescein or other ad- 
sorption indicator. 

(Chloride) (Mohr method, for small amounts) Titrated in neutral solution 
with standard AgNO 3 using K 2 CrO 4 indicator. 

(Free chlorine) Excess KI added and the liberated I 2 titrated with standard 
Na 2 S 2 3 (C1 2 + 21- -+ I 2 + 2C1-). 

(Hypochlorite) Excess KI added in the presence of acid and the liberated 
I 2 titrated with standard Na 2 S 2 O 3 (OC1- + 21- + 211+ -> I 2 + Cl~ + H 2 O). 

(Hypochlorite) Titrated with standard Na 3 AsO 3 using KI + starch as 
outside indicator (OC1~ + AsOr - Cl~ + AsO 4 s ). 

(Chlorate) Reduced to chloride with Zn, FeSO 4 , or H 2 SO 3 and chloride 
determined gravimetrically as AgCl. 

Chromium 

Chromic ions precipitated with NH 4 OH as Cr(OH) 3j ignited, and weighed 
as Cr 2 O 3 . 



COMMON ANALYTICAL DETERMINATIONS 301 

Chromate precipitated as BaCrO 4 from neutral or buffered acid solution, 
ignited gently, and weighed as such. 

Dichromate reduced with measured amount of FeSO 4 and the excess ferrous 
titrated with standard KMnO 4 , K 2 Cr 2 O 7 , or Ce(SO 4 ) 2 (Cr 2 7 - + 6Fe++ + 
14H+ -> 2Cr+" + + 6Fe+++ + 7H 2 O). 

Dichromate reduced with excess KI and the liberated I 2 titrated with 
standard Na 2 S 2 O 3 (CriOr + 6I~ + 14H+ -> 2Cr+++ + 3I 2 + 7H 2 O). 



Cobalt 

Electrolytically deposited as Co from ammoniacal solution. 

Precipitated with alpha-nitroso-beta-naphthol as Co[CioHeO(NO)]3 and the 
precipitate (1) ignited in O 2 to Co 3 O 4 and weighed, or (2) ignited in H 2 to Co 
and weighed. 

Copper 

Electrolytically deposited as Cu. 

Precipitated with H 2 SO 3 + KCNS and weighed as CuCNS (2Cu++ + 
H 2 SO 3 + 2CNS- + H 2 O - 2CuCNS + SO 4 ~ + 4H+). 

Excess KI added and the liberated I 2 titrated with standard Na 2 S 2 O| 
(2Cu++ + 41- - CuJ, + I 2 ). 

Ammoniacal solution titrated with standard KCN to the point of decoloriza- 
tion (2Cu(NH 3 ) 4 ++ + 7CN- + H 2 O -> 2Cu(CN)r + CNO~ + 6NH 3 + 
2NH 4 + ). 

Precipitated as CuCNS (see above) and the precipitate titrated with 
standard KIO 3 forming I 2 which gives a violet color with chloroform 
(lOCuCNS + 14TO 3 - + 14H+ -+10CU++ + 10S<V + 7I 2 + 10HCN + 2TT 2 O). 
Titration continued to disappearance of this color (2T 2 + IGjT + 5C1~ + 
6H + - 5IC1 + 3H 2 O). Net reaction: 4CuCNS + 7IO 8 - + 14H+ + 7C1~ -^ 
4SOr + 7IC1 + 4IICN + 5H 2 O. 



Cyanide 
(See under Nitrogen.) 

Fluorine 

Precipitated as CaF 2 , ignited, and weighed as such. 

Precipitated as PbFCl from acid solution and weighed as such. 

Titrated with standard Th(NO 3 ) 4 using zirconium-alizarin indicator 
(4F~ + Th+++ -> ThF 4 ). 

Evolved as SiF 4 by action with quartz and concentrated H 2 SO 4 , the gas 
absorbed in water and the solution titrated with standard NaOH using 
phenolphthalein (4IIF + SiO 2 -> SiF 4 + 2IT 2 O; 3SiF 4 + 3H 2 -> 2H 2 SiF 6 + 
H 2 Si0 3 ; H 2 SiF 6 + 6OH~ -> 6F~ + H 2 SiO 3 + 3H 2 O). 



Gold 

Chemically or electrolytically reduced to Au and weighed as such. 
Reduced to metal by measured amount of reducing agent (e.g., oxalate) 
and excess titrated with standard KMnO 4 . 



302 CALCULATIONS OF ANALYTICAL CHEMISTRY 

Hydrogen 

Volatilized as water and loss in weight of sample determined. 

Volatilized as water and measured by gain in weight of absorbing agent 
(e.g., CaCl 2 ). 

(Gas analysis) Absorbed on Pd sponge and loss in volume of gas mixture 
determined. Or volume change measured before and after combustion with O 2 . 

Iodine 

(Iodide) Precipitated as Agl and weighed as such. 

(Iodide) Precipitated as PdI 2 and weighed as such (Br~ and Cl~ not 
precipitated). 

(Iodide) (Volhard method) Measured amount of AgNO 8 added and the 
excess Ag+ titrated with standard KCNS (Ag+ + CNS~ -* AgCNS). 

(Iodide) Excess Fe 2 (SO 4 ) 3 added, liberated I 2 caught in KI solution and 
titrated with standard Na^Oa (2I~ + 2Fe 4++ -> I 2 + 2Fe++) (Br~ and Cl~ 
not affected). 

(Iodide) Excess KIO S added in presence of acid and liberated I 2 boiled out. 
Excess I(V determined in cooled solution by adding KI and titrating the 
liberated I 2 with standard Na^S-A (5I~ + ICV + 6H+ -> 3I 2 + 3H 2 O). 

(Iodide) Titrated with standard KIO 8 in presence of concentrated HC1 
using chloroform as indicator. I 2 is first liberated and colors CHC1 3 violet. 
Color fades away at end point. Net reaction: 2I~ + IO 3 ~ + 3C1~ + 6H+ - 
3IC1 + 3H 2 0. 

(Iodide) Titrated directly with standard AgNO 3 using eosin adsorption 
indicator. 

(lodate) Excess KI added and the liberated I 2 titrated with standard 
Na*S 2 O 8 (IQr + 51- + 6H+ -> 3I 2 + 3TI 2 O). 

(Free iodine) Titrated with standard Na 2 S 2 O 3 using starch indicator 
(I 2 + 2S 2 Or -* 21- + S 4 6 -). 

(Free iodine) Titrated with standard NasAsOa using starch indicator in 
solution kept nearly neutral with excess NaHCQe (12 + AsOs 13 + 2HCO 8 - -> 
AsO 4 " + 21- + 2CO 2 + H 2 0). 

Iron 

Precipitated as Fe(OH) 3 with NH 4 OH or NaOH, ignited, and weighed as 
Fe 2 8 . 

Precipitated with cupferron as (CeHsNONO^Fe from acid solution, ignited, 
and weighed as Fe 2 O 3 . 

Ferrous titrated with standard KMnO 4 , K 2 Cr 2 7 , orCe(SO 4 ) 2 (e.g., 5Fe++ + 
MnO 4 - + 8H+ -> SFe+++ + Mn^ + 4TI 2 O). 

Ferric treated with large excess KI and liberated I 2 titrated with standard 
Na 2 S 2 O 3 (2Fe + ^ + 21" -> I 2 + 2Fe^). 

Ferric titrated with standard TiCl 3 solution using KCNS indicator 



Lead 

Precipitated as PbSO 4 or PbCrO 4 or PbMoO 4 and weighed as such. 
Electrolytically oxidized and deposited as PbO 2 , and weighed as such 
(Pb++ + 2H 2 - PbO 2 



COMMON ANALYTICAL DETERMINATIONS 303 

Titrated with standard (NH 4 ) 2 MoO4 using tannin as outside indicator 

+ MoOr -> PbMoO 4 ). 
Precipitated as PbCrO 4 , the precipitate dissolved in acid and the Cr 2 07~ 
determined volumetrically as under Chromium above. 

Magnesium 
Precipitated from ammoniacal solution as MgNH 4 P0 4 , ignited, and 

weighed as Mg 2 P2O7. 

Manganese 

Manganous ions oxidized by KC1O 3 or KBrO 3 to Mn0 2 . Precipitate ignited 
in air and weighed as Mn 3 O4. 

Precipitated as MnNH 4 PO4, ignited, and weighed as Mn 2 P 2 Or. 

(Bismuthate method) Oxidized with NaBiO 3 or BiO 2 to permanganate. 
Measured amount of FeSO 4 added, and excess ferrous titrated with standard 
KMnO 4 (2Mn++ + 5NaBiO 8 + 14H+ -> 2MiiO 4 - + 5Bi+++ + 5Na+ + 7TI 2 O). 

(Ford-Williams method) Oxidized with KC1O 3 in presence of concentrated 
HNO 3 to MnO 2 . Measured amount of FeSO 4 added, and the exceiss ferrous 
titrated with standard KMn0 4 (MnO 2 + 2Fe++ + 411+ -> Mn+ + + 2Fe+++ + 
2H 2 O). 

(Persulfate method) Oxidized with (NH 4 ) 2 S 2 O 8 (+AgNO 3 ) to perman- 
ganate, and then titrated with standard Na 3 AsO 3 to indefinite valence of 3+. 
Arsenite standardized against similar sample containing known Mn. 

(Volhard method) Manganous ions titrated directly with standard KMnO 4 
in solution kept neutral with ZnO (3Mn++ + 2Mn0 4 ~ + 2ZnO - 5MnO 2 + 

2Zn++ + 2II 2 O). 

Mercury * 

Precipitated as HgS and weighed as such. 

Electrolytically precipitated as Ilg and weighed as such. 

Titrated with standard KCNS using ferric alum indicator [Tig"*" 1 " + 
2CNS- -> Hg(CNS) 2 ]. 

Molybdenum 

Precipitated as PbMoO 4 or Ag 2 MoO 4 and weighed as such. 

Precipitated as Hg 2 MoO 4 or as MoS 3 , ignited, and weighed as MoO 3 . 

Reduced with Zn and passed directly into ferric alum. The reduced iron 
then titrated with standard KMiiO 4 (2MoOr + 3Zn + 1611+ - 2Mo f f + + 
3Zn++ + 8H 2 O; 5Mo+++ + 3MnO 4 " + 8II 2 O -> 5MoO 4 - + 3Mn^ + 16II+). 

Nickel 

Precipitated with dimethyl glyoxime as [(CH 3 ) 2 .CNOH.CNO] 2 Ni and 
weighed as such. 

Electrolytically precipitated as Ni and weighed as such. 

Measured amount of KCN added to ammoniacal solution and the excess 
CN- titrated with standard AgNO 3 using KI indicator [Ni(NH 3 )++ + 
4CN- -> Ni(CN) 4 - .+ 6NH 3 ; 2CN~ + Ag + -^ Ag(CN) 2 -]. 

Nitrogen 

(Organic nitrogen) (Kjeldahl method) Converted by digestion with con- 
centrated H2S0 4 + catalyst to NH 4 HSO 4 . Excess NaOH is then added, the 



304 CALCULATIONS OF ANALYTICAL CHEMISTRY 

liberated NH 8 distilled into measured amount of acid, and the excess acid 
titrated with standard NaOH using methyl red indicator. 

(Ammonium) Excess NaOH added, the liberated NH 3 distilled into 
measured amount of acid, and the excess acid titrated with standard NaOH 
using methyl red indicator. 

(Ammonium) Precipitated as (NH 4 ) 2 PtCl 6 and weighed as such, or ignited 
toPt. 

(Nitrate) Reduced to NH 4 + with Zn or with Devarda alloy; then by 
Kjeldahl method above. 

(Nitrite) Titrated with standard KMnO 4 (5NO 2 ~ + 2Mn0 4 ~ + 6H+ -> 
5NO 8 - + 2Mn++ + 3H 2 O). 

(Cyanide) (Volhard method) Measured amount of AgNO 8 added and the 
excess Ag+ titrated with standard KCNS using ferric alum indicator (2CN~ + 
2Ag+ -> Ag 2 (CN) 2 ). 

(Cyanide) (Liebig method) Titrated with standard AgNO 3 to faint turbidity 
of Ag 2 (CN) 2 [2CN- + Ag+ -> Ag(CN)r]. 

(Gas analysis) Volume of residual nitrogen measured after absorbing 
other gases. 

Oxalate 

(See under Carbon.) 

Oxygen 

(Gas analysis) Volume of gas mixture determined before and after ab- 
sorbing in alkaline pyrogallol. 

Phosphorus 

Precipitated as MgNH 4 PO 4 , ignited, and weighed as Mg 2 P 2 O 7 . 

(Iron and steel) Precipitated as (NH 4 ) 8 PO 4 .12MoO 3 and weighed as such, 
or ignited and weighed as P 2 O5.24MoO 3 . 

(Iron and steel) (Ferric alum metlwd) Precipitated as (NH 4 ) 3 P0 4 .12Mo0 8 , 
dissolved, the Mo reduced with Zn in a Jones reductor and passed directly 
into excess ferric alum. The reduced iron then titrated with standard 
KMnO 4 (2MoOr + 3Zn + 16H+ -> 2Mo ++ + + 3Zn++ + 8II 2 O; MO+++ + 
3Fe+++ + 4H 2 O -> MoOr + 3Fe++ + 8H+). 

(Iron and steel) (Blair method) As in the preceding method except that the 
reduced solution is caught in an open flask where slight oxidation by the air 
occurs. The Mo, now having an average valence corresponding to the oxide 
Mo 24 37 , is titrated to MoO 4 " with standard KMnO 4 . 

(Iron and steel) (Alkalimetric method) Precipitated as (NH 4 ) 8 PO 4 .12MoO 3 , 
dissolved in a measured amount of standard NaOH and the excess alkali 
titrated with standard HNO 3 using phenolphthalein indicator. Net reaction: 
(NH 4 ) 3 PO 4 .12MoO 8 + 23OH- - 12MoO 4 - + HP0 4 " + 3NH 4 + + 11H 2 0. 

Platinum 

Precipitated as K 2 PtCl 6 and weighed as such. 

Precipitated as (NH 4 ) 2 PtCl6 and weighed as such, or] ignited to Pt and 
weighed. 

Electrolytically reduced to Pt and weighed as such. 



COMMON ANALYTICAL DETERMINATIONS 305 

Potassium 

Precipitated as K 2 PtCl6 and weighed as such, or the precipitate reduced 
to Ft and weighed. 

Precipitated as KC1O 4 and weighed as such. 
See also under Sodium. 

Selenium 
Reduced by H 2 SO 8 , KI, etc., to Se and weighed as such. 

Silicon 

Precipitated as H 2 SiO 8 , ignited to SiO 2 and weighed. The impure SiO 2 is 
then treated with HF, evaporated, reignited, and impurities are weighed. 
Loss in weight = SiO 2 (SiO 2 + 4HF -> SiF 4 + 2H 2 O). 

Silver 

Precipitated as AgCl and weighed as such. 

(Volhard method) Titrated with standard KCNS using ferric alum indi- 
cator (Ag+ + CNS- -> AgCNS). 

Titrated with standard NaCl using fluoresceiii as an adsorption indicator. 

Sodium 

(Silicates) (J. Lawrence Smith method) Silicate decomposed by heating 
with CaCO 3 + NH 4 C1. Leached with water, Ca++ removed, filtrate evapo- 
rated, and residue ignited. NaCl + KC1 weighed. K then determined as 
KC1O 4 or K 2 PtCl e . Na determined by difference. 

(Small amounts) Precipitated as NaZn(UO 2 )3(C2H 3 O 2 )9.6H 2 O or as 
NaMg(UO 2 )3(C 2 H 3 O 2 )9.6^2H 2 O and weighed as such 

Strontium 

Precipitated as SrS0 4 and weighed as such. 

Sulfur 

(Sulfate) Precipitated as BaSO 4 and weighed as such. 

(Sulfate) (Hinman method) Excess acid solution of BaCrO 4 added (SO 4 ~ + 
Ba++CrO 4 ~ - BaSO 4 + CrOr). Excess NH 4 OH added to precipitate excess 
BaCrO 4 . Combined BaSO 4 + BaCrO 4 filtered. Filtrate acidified, treated 
with excess KI, and liberated I 2 titrated with standard Na 2 S 2 O 8 (Cr 2 O 7 ~ + 
61- + 14H+ -2Cr ++ + + 3I 2 + 7H 2 O). In the titration each Cr 2 O 7 - is 
equivalent to 2SO 4 ~. 

(Sulfate) Precipitated with benzidine hydrochloride giving Ci 2 H 8 (NH 2 ) 2 .- 
H 2 SO 4 . Suspension of precipitate titrated with standard NaOH which acts 
only on the H 2 SO 4 . 

(Sulfide) (Evolution method for alloys) Evolved as H 2 S by action of HC1 and 
caught in ammoniacal solution of ZnSO 4 . The solution is acidified and the 
H 2 S titrated with standard I 2 or standard KIOs + KI using starch indicator 



(Sulfide) Oxidized to sulfate, precipitated as BaSO 4 and weighed. 



306 CALCULATIONS OF ANALYTICAL CHEMISTRY 

(Sulfite) Titrated with standard I 2 using starch indicator. (SOs~ + 
H 2 O + I* -> SOr + 21- + 2H+0. 

(Persulfate) Measured amount of FeS0 4 added and excess ferrous titrated 
with standard KMnO 4 (S 2 O 8 ~ + 2Fe++ -> 2SO 4 - + 2Fe^+). 

(Thiosulfate) Titrated with standard I 2 (2S 2 3 ~ + 1 2 - S 4 O 6 " + 21 -). 

(Thiocyanate) Measured amount of AgNO 8 added and excess Ag+ titrated 
with standard KCNS (CNS- + Ag + - AgCNS). 

Thorium 
Precipitated as Th(C 2 O 4 ) 2 , ignited, and weighed as ThO 2 . 

Tin 

Precipitated as H 2 SnO 3 by hydrolysis, ignited, and weighed as SnO 2 . 
Precipitated as SnS 2 , ignited, and weighed as Sn() 2 . 

(In alloys) Alloy treated with 1INO 3 leaving H 2 SnO 3 as residue. This is 
ignited to SnO 2 and weighed. Antimony must be removed. (3Sn + 4NO 3 ~ + 
1I 2 O -> 3H 2 SnQ 3 + 4NO). 



Titrated from 2 to 4 in cold IIC1 solution in current of CO 2 with standard 
I 2 (Sn-* + + I 2 + 6C1- - SnClT + 21"). 

Titrated to yellow color with standard FeCl 3 (Sn++ + 2FeClr Sn-Cle~ + 
2Fe + + + 2C1-). 

Titanium 

Precipitated as Ti(OH) 4 , ignited, and weighed as Ti0 2 . 

Precipitated with cupferron, ignited, and weighed as TiO 2 . 

Reduced by Zn (but not reduced by SnCl 2 ) to Ti+++ and titrated with 
standard KMnO 4 , or passed from zinc reductor into ferric alum and the re- 
duced iron titrated with standard KMnO 4 (2Ti++++ + Zn - 2Ti+++ + 
Zn++; 5Ti +++ + MnO 4 ~ + 8H+ -> 5Ti ++ ^ + Mn++ + 4H 2 O). 

Reduced to Ti +++ and titrated with standard ferric alum using NH 4 CNS 
indicator (Ti+++ + Fe^ + -> Ti ++++ + Fe++). 



Tungsten 

Precipitated with acid as H 2 W0 4 or with cinchonine as cinchonine tungstate, 
ignited, and weighed as WO 3 . 

Uranium 
Precipitated with NH 4 OH as (NH 4 ) 2 U 2 O 7 , ignited in air, and weighed as 

U 3 8 . 

Precipitated as UO 2 NII 4 PO 4 , ignited, and weighed as (UO 2 ) 2 P 2 O 7 . 

Reduced with Zn and titrated back with standard KMnO 4 (UO 2 + + + Zn 4- 
4H+ -* U++++ + Zn ++ + 2H 2 O; 5U++++ + 2Mn0 4 - + 2H 2 O -> 5UO 2 ++ + 
2Mn +4 - + 4H+). 

Vanadium 

Precipitated as HgVO 3 , ignited, and weighed as V 2 O 6 . 
Precipitated as Pb(V0 8 )2, fumed with H 2 S0 4 , filtered, ignited, and weighed 
as V 2 O 6 . 



COMMON ANALYTICAL DETERMINATIONS 307 

Reduced from 5 to 4 by 862 or H 2 S and titrated back with standard KMn0 4 
(2V0 8 - + S0 2 + 4H+ -> 2VO++ + S0 4 " + 2H 2 0; 5VO++ + MnO 4 ~ + 6H 2 -> 
5V0 8 ~ + Mn++ 



Zinc 

Precipitated as ZnS, ignited, and weighed as ZnO. 

Precipitated as ZnNH 4 PO 4 and weighed as such or ignited to Zn 2 P207 
and weighed. 

Titrated with standard K4Fe(CN) 6 using FeS0 4 as internal indicator or 
UO 2 (NO 8 ) 2 as external indicator. The net equation is: 3Zn++ + 2Fe(CN) 6 li + 
2K+->K 2 Zn 3 [Fe(CN) 6 ] 2 . 

Zirconium 

Precipitated as Zr(OH) 4 , ignited, and weighed as ZrO 2 . 
Precipitated with cupferron or phenyl-arsonic acid, ignited, and weighed 
QS ZrO 2 . 

Precipitated with H 2 Se0 8 as Zr()ScO3 t ignited, and weighed as ZrO 2 . 
Precipitated as Zr(HP0 4 ) 2 , ignited, and weighed as ZrP 2 7 . 



PART VII 

PROBLEMS ON SPECIFIC GROUPS 
AND DETERMINATIONS 

A. QUALITATIVE ANALYSIS 

Silver Group 
(See also Probs. 54, 66, 109, 126, 153, 155, 174.) 

837. Silver chloride dissolves in NH 4 OH according to the equation: 
AgCl + 2NII 4 OH -> Ag(NH 3 ) 2 + + Cl- + 2H 2 O. Calculate (a) number of 
gram-moles of NII 4 OH equivalent to 1 F.W. of AgCl, (b) number of grams of 
NH 3 equivalent to 1 F.W. of AgCl, (c) number of gram-ions of Cl"" produced 
by dissolving 1 F.W. of AgCl. 

838. Assuming 20 drops to equal a milliliter, how many drops of 2.0 formal 
NH 4 C1 solution would be required to precipitate the silver from a solution 
containing 100 mg. of AgNO 3 ? How many millimoles and how many milli- 
grams of NII 4 C1 does each milliliter of the reagent contain? How many milli- 
liters of 5.0 normal NH 4 C1 should be taken in order to prepare 500 ml. of the 
2.0 formal solution by dilution with water? 

839. A neutral solution containing 0.0170 gram of dissolved AgN0 3 is 
treated with an aqueous solution of 0.120 millimole of HC1 and the precipi- 
tated AgCl is filtered off. (a) How many gram-ions of Cl" and how many 
grams of Cl~ are present in the filtrate? (6) How many milliliters of NH 4 OH 
(sp. gr. 0.96 containing 9.9 per cent NH 3 by weight) would be required to 
neutralize the acid in the filtrate? (c) How many milliliters of 2.0 N NH 4 OH 
would be required to dissolve the AgCl precipitate? 

840. Complete and balance the following equation: Hg 2 Cl2 + NH 4 OH > 
Hg + HgNTI 2 Cl + . . . . What oxidation numbers does mercury show in 
this reaction? How many grams of mercurous chloride would give 0.0100 F.W. 
of the amido compound by this reaction? How many gram atoms of free mer- 
cury would be formed at the same time, and how many grams of ammonium 
chloride could be obtained by filtering and evaporating the filtrate to dryness? 
How many milliliters of ammonium hydroxide (sp. gr. 0.970, containing 7.31 
per cent NH 3 by weight) would be required in the reaction? 

841. What is the solubility product of lead chloride, PbCl 2 , if 550 milligrams 
dissolve in 50.0 ml.? How many milligrams of chloride ions must be present 
in 3.00 ml. of a solution containing 0.100 millimole of Pb(N0 8 ) 2 in order to 
give a precipitate of lead chloride in the silver group? 

309 



310 CALCULATIONS OF ANALYTICAL CHEMISTRY 

842. To a solution containing 50.0 mg. of Ag + and 50.0 mg. of Pb++ are 
added sufficient NH 4 C1 solution to give precipitates of AgCl and PbCl 2 and 
make the surrounding solution half normal in chloride ions. If the volume of 
the solution is 30.0 ml and the solubility products of AgCl and PbCl 2 are 
1.0 X 10~ 10 and 2.4 X 10~ 4 , respectively, how many milligrams of silver and of 
lead will remain unprecipitated? How many milliliters of boiling water are 
required to dissolve the precipitated lead chloride if its solubility at 100C. is 
0.120 F.W. per liter of water? How many milligrams of silver chloride would 
dissolve by this treatment if its solubility at 100C. is 0.150 millimoles per liter 
of water? 

Hydrogen Sulfide Group 

(See also probs. 55, 57, 67, 106, 108, 110, 123, 128, 162, 163, 167, 171, 172, 
176.) 

843. If precipitation of suifides is carried out in a solution 0.30 N in hy- 
drogen ions, what is the pH value of the solution? What is the hydroxyl-ion 
concentration? 

844. A solution contains dilute HN0 3 and 0.0485 gram of dissolved 
Bi(NO 3 ) 3 .5H 2 O. The bismuth is precipitated by H 2 S as follows: 2Bi+++ + 
3H 2 S Bi 2 S 3 + 6H+. Calculate (a) number of formula weights of bismuth 
sulfide produced, (6) number of grams and number of milliliters (standard 
conditions) of H 2 S required, (c) increase in the number of gram-ions of H + 
accompanying the precipitation. 

846. The bismuth sulfide of the preceding problem is dissolved in HNOa 
according to the ionic equation: Ri 2 S 3 + 2NO 3 ~ + 8H+ - 2Bi+++ + 2NO + 
3S + 4H 2 O; 2NO + O 2 (air) ~-> 2NO 2 . Calculate (a) number of formula 
weights of Bi(NO 3 ) 3 .5H 2 O obtainable from the resulting solution, (fc) number 
of formula weights of HNO 3 required; (c) number of milliliters of HN0 3 
(Sp. gr. 1.13, containing 21.8 per cent HN0 3 by weight) required, (d) number 
of gram-atoms of sulfur produced, (e) number of millimoles of NO gas formed; 
(/) number of grams of NO 2 subsequently produced, (g) percentage of nitrogen 
in the NO 2 gas. 

846. How many milliliters of H 2 S0 4 (sp. gr. 1.14, containing 19.6 per cent 
H 2 SO4 by weight) are theoretically required to precipitate 10.0 milligrams of 
Pb++ ? What is the normality of the above sulfuric acid and what is its com- 
position in terms of percentage of combined 80s? How many formula weights 
of NH 4 C2H 3 02 would be required to dissove the resulting precipitate of lead 
sulfate? If the latter solution were diluted to 50.0 ml., what would be its 
normality in terms of lead acetate and its molarity in terms of ammonium 
sulfate? 

847. When H 2 S is passed into an acid solution containing 0.10 F.W. of 
K 2 Cr 2 O 7 , the following reaction takes place: K 2 Cr 2 O 7 + 3H 2 S + 8HC1 -*> 
2CrCU + 2KC1 + 3S + 7H 2 0. (a) Write this equation in ionic form ob- 
serving the usual conventions, (b) Express the reaction as the difference 
between two half-cell reactions, (c) How many formula weights of chromic 
chloride and how many grams of potassium chloride could be obtained by 



PROBLEMS 311 

evaporating the resulting solution to dryness? (d) How many gram-atoms 
of sulfur are formed? (e) How many milUmoles of H 2 S are oxidized? (/) How 
many milliliters (standard conditions) of H 2 S are oxidized? (g) If the initial 
solution has a volume of 100 ml. and is 0.30 N in HC1, what is the acid nor- 
mality of the solution after the above reaction has taken place, assuming no 
appreciable change in volume? 

848. In the precipitation of the copper-tin groups from acid solution with 
H 2 S, oxidizing agents like permanganate ions and ferric ions are reduced by 
the H 2 S and the latter is oxidized to free sulfur. Write each of these two 
redox reactions as the difference between two half-cell reactions, and calculate 
the number of millimoles and the number of milligrams of KMnO 4 and of FeCU 
thus reduced by 10.0 ml. of 11 2 S gas (measured under standard conditions). 

849. Balance the following equation and express it as the difference between 
two half-cell reactions: HgS + C10 3 ~ + Cl~ -> HgClr + SOr. Assume that 
0.233 gram of HgS is dissolved according to this equation and the solution 
diluted to 50.0 ml. If excess chloride ions are present in sufficient amount to 
make the chloride-ion concentration 0.510 molar, how many milligrams of 
mercury are present as simple Hg H f ions? (Dissociation constant of HgClr = 
1.0 X 10- 16 ). 

860. A solution containing 0.010 F.W. of CdS0 4 and 0.010 F.W. of CuSO 4 is 
made ammoniacal and treated with excess KCN, forming Cd(CN) 4 ~ and 
Cu(CN)a~. The resulting solution is 0.80 molar in CN~ ions. What is the 
molar concentration of Cd++ and of Cu+ in the solution if the dissociation 
constant of Cd(CN) 4 - is 1.4 X 10~ 17 and that of Cu(CN) 3 - is 5 X 10~ 28 ? 
Are these values consistent with what happens when II 2 S is passed into the 
complex cyanide solution? 

851. Mercuric sulfide dissolves in a solution of Na2S + NaOII (HgS + 
S- -> HgS 2 -) but not in a solution of (NII 4 ) 2 S + NH 4 OH. This is due to the 
difference in the degree of hydrolysis of the sulfide ion: S" + 1^0 IIS~ + 
OH". Calculate the numerical value of the mass action hydrolysis constant 
[HS~~][OH~]/[S~] by combining appropriate ionization constants given in 
Sees. 27 and 3L 

852. A solution containing 30 milligram-atoms of bismuth as Bi 4f " f is 
treated with I1C1 in sufficient amount to make the hydrogen-ion concentration 
2 molar and the chloride-ion concentration 2 molar. Most of the bismuth is 
converted to BiCl 4 ~. The solution has a volume of 15 ml. and is made 0.10 
molar in I^S. From the ionization constant of H2S calculate the concentration 
of S" and from the dissociation constant of BiCU~ calculate the concentration 
of Bi+++. Calculate the value of [Bi+++] 2 [S-] 3 and predict from the solubility 
product of Bi 2 Ss whether a precipitation of the sulfide is to be expected. 

Ammonium Sulfide Group 
(See also Probs. 78, 120, 127, 140, 141, 154, 160, 161, 168, 169, 170, 173, 185.) 

853. A solution contains 0.286 gram of dissolved Fe 2 (S0 4 ) 3 .9H 2 O in 500 ml. 
of solution, (a) What is the normality of the solution as a ferric salt? 



312 CALCULATIONS OF ANALYTICAL CHEMISTRY 

(b) What is its normality as a sulfate? (c) How many milliliters of 3.00 N 
NH 4 OH are required to precipitate all the iron as Fe(OH) 3 ? (d) How many 
milliliters of barium chloride solution containing ^ millimole of BaCl 2 .2H 2 O 
per milliliter are required to precipitate the sulfate? 

854. From the solubility products of Mg(OH) 2 and Fe(OH) 3 calculate the 
number of grams of ferric ions that can remain dissolved in 100 ml. of a solution 
of such alkalinity that 243 mg. of Mg++ will just fail to precipitate as Mg(OIT) 2 . 

866. What is the molarity of a solution of CrCl 3 which is 0.10 normal as a 
salt? How many milliequivalents of the salt are present in each milliliter? 
If the chromium is oxidized to dichromate and the volume is twice as great 
as before, what would be the normality as a sodium salt of the Na 2 Cr 2 O 7 
present? How many inillimoles of FeSO 4 .7H 2 O would be required to reduce 
this dichromate to chromic ions in the presence of acid? Write the ionic equa- 
tion as the difference between two half-cell reactions. 

866. How many milliliters of H 2 SO 4 (sp. gr. 1.20) are theoretically required 
to dissolve 0.010 F.W. of A1(01I) 3 ? What is the normality and what is the 
molarity of the acid? 

867. How many millimoles of HNO 3 are required to dissolve 0.020 gram- 
atom of metallic nickel (HNO 3 reduced to NO)? How many grams of NiS 
could be precipitated from the solution? Show from appropriate mass-action 
constants why one would not expect precipitate to form from 100 ml. of the 
above solution if it has been made 0.30 normal in H+ and 0.10 molar in H 2 S. 
How high a molar concentration of sulfide ion would be necessary before NiS 
would start to precipitate? 

858. Starting with 0.010 gram-ion of Zn + +, how many millimoles of NH 3 
are theoretically required to form the ammonio complex ion? If three times 
this amount of ammonia are used and the total volume is 250 ml., what is the 
molar concentration of Zn+ + ? 

Alkaline Earth and Alkali Groups 
(See also Probs. 124, 135, 142, 158, 177.) 

869. A solution is 0.030 formal in Sr++ ions. What value must the chromate- 
ion concentration be in order for SrCrO 4 to start precipitating (Ks.p. for 
SrCrO 4 = 3.0 X 10~ 6 )? How many grams of Ba++ could remain dissolved 
in each milliliter of such a solution (K s .p. for BaCrO 4 = 3.0 X 10~ 10 )? The 
mass-action constant for the equilibrium 2CrO 4 =B + 2H+ <= Cr 2 O 7 ~ + II 2 O is 
4.2 X 10 14 . If the above solution contains sufficient HC 2 H 3 02 and NH 4 C 2 H 3 O2 
to give a pH value of 5, what would be the dichromate-ion concentration? 

860. How many grams of CaCl 2 should be taken to prepare 100 ml. of 0.20 N 
solution? How many millimoles of H 2 C 2 O 4 .2H 2 O would be required to pre- 
cipitate all the calcium? How many gram atoms of magnesium could theo- 
retically be held as Mg(C 2 O 4 ) 2 ~ by this amount of oxalic acid? 

861. How many milliliters of a solution of K 2 Cr 2 Oy which is 0.10 normal as a 
potassium salt would be required to precipitate 0.010 gram-atom of barium 
as BaCr0 4 ? 



PROBLEMS 313 

862. How many grams of NH 3 would be liberated from 1.00 gram-equivalent 
weight of (NH 4 ) 2 SO 4 by the action of NaOH? If this NH 3 were absorbed in 
water and diluted to one liter, what would be the normality and the approxi- 
mate specific gravity of the solution? How many milliliters of 3.40 M H 2 SO4 

would it neutralize? 

Anion Groups 

(See also Probs. 107, 125, 130, 156, 157, 159, 165, 178, 179, 182.) 

863. Complete and balance the following: NO 3 ~ + A 1 + OH- -> NH 8 + 
A1O 2 ". Also write it as the difference between two half-cell reactions. If 250 
mg. of NaNO 3 are reduced as above with excess aluminum in the presence of 
NaOH, how many milliliters of N/2 II 2 SO 4 would be required to neutralize the 
NH 8 liberated? How many gram-atoms of aluminum are theoretically re- 
quired for the reduction? 

864. If 100 milligrams of sodium oxalate Na 2 C 2 O 4 , are heated with concen- 
trated H 2 SO 4 , what volume of mixed gases would be obtained when measured 
over water at 753 mm. pressure and 25C.? (Na 2 C 2 O 4 + H 2 SO 4 > Na 2 SO 4 + 
(JO _j_ CO 2 + II 2 O.) What volume of gas would be obtained (under the same 
conditions of temperature and pressure as above) if the 100 mg. of sodium 
oxalate were treated with excess KMnO 4 in the presence of dilute H 2 SO 4 ? 

B. QUANTITATIVE ANALYSIS 

Water 
(See also Probs. 259, 360, 361, 367, 368, 369, 372, 373, 374, 375, 377.) 

866. A manufacturer purchased 130 tons of material at 0.20 cent per pound 
per cent A on a guarantee of 10.00 per cent V A. The material was shipped in 
cars, and on arrival the manufacturer had it analyzed. The chemist reported 
10.46 per cent A but neglected to state that he had dried the sample at 100O. 
The manufacturer paid on a 10 per cent basis, figuring he had made money. 
In reality, he lost $520. What was the percentage of moisture in the material? 

866. Ten tons of Na 2 SO 3 containing 6.30 per cent moisture were purchased 
at the market price. During storage, 10.0 per cent of the sodium sulfite was 
oxidized to sodium sulfate. The salt when sold contained 3.20 per cent of 
its weight of water. The salt was sold as C.P. Na 2 SO 3 at the same price as 
purchased. Calculate the gain or loss in the transaction figuring the market price 

as 3.25 cents per pound. 

Sodium. Potassium 

(See also Probs. 283, 306, 307, 313, 314, 322, 539, 541, 542, 543, 723, 724, 
725, 730, 731, 732, 734, 1028.) 

867. If in the analysis of a silicate a mixture of pure NaCI and KC1 weighing 
0.2500 gram was obtained from a sample weighing 0.7500 gram and if this 
mixture of chlorides contained 50.00 per cent chlorine, compute the percentage 
of K 2 O in the silicate. What weight of potassium perchlorate would have been 
obtained if the chlorides had been analyzed by the perchlorate method? 

868. In the J. Lawrence Smith method for potassium using a 0.5000-gram 
sample of mineral, the analyst fails to expel all the ammonium chloride from 



314 CALCULATIONS OF ANALYTICAL CHEMISTRY 

the NaCl + KC1. The insoluble precipitate with chloroplatinic acid weighs 
0.08921 gram. On ignition the weight is changed to 0.05969 gram. What 
would be the weight if the ignited precipitate were washed with water and 
dried, and what is the percentage of K 2 O in the mineral? 

869. With a 0.5000-gram sample of pure NaKCaSi 2 O 6 (mol. wt. = 254.3), 
what would be the financial saving of the perchloric acid method over the 
chloroplatinic acid method with 3.00 N perchloric acid at $18 per liter and 
platinum at $6 per gram? Assume these reagents to be added in sufficient 
amounts to react with both the sodium and potassium in the sample and that 
80 per cent of the platinum is recovered. Neglect all other costs. 

870. Caustic potash is to be produced by the electrolysis of a solution of 
potassium chloride. A solution containing 100 grams of KOH per liter is 
required. An average current of 900 amperes is used; and, at the end of 5.00 
hours, 102 liters of caustic potash, 1.520 N as an alkali, have been produced. 
How much longer must the electrolysis be continued in order to produce the 
desired concentration, and what is the current efficiency at the cathode? 

871. How many grams of Pt (dissolved in aqua regia) and how many milli- 
liters of HC1O 4 (3.000 N as an acid) would theoretically be required to pre- 
cipitate the potassium from 0.5000 gram of K 3 PO 4 without allowing for the 
customary excess of reagent? How much would the former precipitate weigh 
after ignition? 

872. When an electric current is passed through a solution of sodium 
chloride, metallic mercury being used as a cathode, metallic sodium is liberated 
and dissolves in the mercury, forming a hard gray compound which shows the 
properties of an alloy and is called an amalgam. Sodium amalgam is used 
extensively as a reducing agent and contains a compound of the formula 
NaHg 2 . 

Five grains of the amalgam are placed in a flask containing about 100 ml. 
of water and allowed to stand with repeated shaking until the evolution of 
gas has entirely ceased. The solution is then titrated with 40.75 ml. of 
0.1067 N of HC1, methyl orange being used us an indicator. Write the 
equation for the reaction of NaHg 2 on H 2 O. Calculate the percentage of Na 
in the sample. 

Ammonium. Ammonia. Nitrogen 

(See also Probs. 250, 258, 280, 520, 523, 524, 525, 532, 544, 1011.) 

873. Ten milliliters of ammonium hydroxide (sp. gr. 0.960) are diluted 
to exactly 100 ml. in a calibrated flask; 10.00 ml. are withdrawn in a pipet, 
made acid with hydrochloric acid, and an excess of chloroplatinic acid is 
added. After evaporation and dilution with alcohol, the insoluble residue is 
dried and found to weigh 1.240 grams. Calculate the percentage of NH 8 by 
weight in the original ammonia sample. 

874. The nitrogen in a half-gram sample of urea, CO(NH 2 )2, is determined 
by the Kjeldahl method. The evolved ammonia is passed into 150 ml. of 
0.1200 N H 2 SO 4 . How many ml. of NaOH solution would be required for the 
excess acid if 1.000 ml. NaOII o 0.00700 gram of hydrated oxalic acid? 



PROBLEMS 315 

875. The nitrogen in 5.000 grams of leather is converted into ammonium 
bisulfate. After the sample has been treated with excess NaOH, the NH 3 
evolved is passed into 90.0 ml. of 0.4990 N acid and the excess acid titrated with 
22.05 ml. of 0.1015 N NaOH. Find the percentage of nitrogen in the leather. 

876. Nitrogen, existing as nitride in a crucible steel, is determined by de- 
composing a 5.00-gram sample with HC1. The resulting NH 4 C1 is decomposed 
with NaOH, and the liberated NH 3 is absorbed in 10.05 ml. of H 2 SO 4 which is 
exactly 0.00990 N as an acid. After absorption, the concentration of the 
II 2 SO 4 is determined by adding an excess of KI and of KIO 8 and titrating with 
standard Na 2 S 2 O3 the I 2 liberated. The Na 2 S 2 O 3 is of such strength that 42.0 
ml. are equivalent to the I 2 liberated from an excess of KI by 20.0 ml. of 0.0258 
N KMnO 4 , and in the above titration 5.14 ml. are used. Calculate the percent- 
age of nitrogen in the steel. 

877. A sample of impure ammonium chloride is dissolved in water, and the 
solution is divided into two equal portions. One portion is made alkaline 
with NaOH, and the liberated ammonia is distilled into 100 ml. of 0.1000 N 
sulfuric acid which is then found to require 43.90 ml. of 0.1320 N NaOH for 
neutralization. The other portion is treated with sodium hypobromite solution 
(2NH 3 + 3OBr~ > 3Br~ + N 2 + 3H 2 0), and the liberated nitrogen is found 
to occupy 51.30 ml. when measured over water at 20C. and 753 mm. pressure. 
If the first method gives correct results, what is the percentage error of the 
gas-volumetric method? 

Silver. Mercury. Gold. Platinum 
(See also Probs. 333, 335, 338, 341, 346, 718.) 

878. If a sample of silver coin weighing 0.2500 gram gives a precipitate of 
AgCl weighing 0.2991 gram, what is the percentage of silver in the coin and 
what volume of 0.05000 N KCNS solution would have been used if the silver 
in the same weight of sample had been determined volumetrically by the 
Volhard method? 

879. Mercury, like silver, forms an insoluble thiocyanate [IIg ++ + 2CNS~~ > 
Hg(CNS) 2 ] and can be determined by titration with standard KCNS. 
How many milliliters of 0.08333 N KCNS would be required to titrate the 
solution of 0.6000 gram of an amalgam consisting of 70.00 per cent Hg and 
30.00 per cent Ag? 

880. What weight of metal can be deposited at the cathode in 20 minutes 
by the electrolysis of (a) HAuCU, and (b) H 2 PtCl 6r using an average current 
strength of 1.50 amperes? What material would you use for the construction 
of the anode? 

Halogens. Cyanide. Thiocyanate. Halogen acids 
(See also Probs. 249, 257, 268, 286, 288, 298, 308, 311, 316, 324, 325, 326, 
327, 328, 329, 330, 331, 695, 713, 720, 721, 722, 726, 735, 736, 737, 738, 740, 
741, 744, 745, 746, 747, 748, 1013, 1014, 1018, 1019, 1027, 1028.) 

881. In the determination of fluoride in a given sample of a salt mixture, 
if 20.00 ml. of NaOH (1.000 ml. o 0.01021 gram potassium acid phthalate) 



316 CALCULATIONS OF ANALYTICAL CHEMISTRY 

were required, what weight of precipitate would be obtained if the same weight 
of sample were analyzed for fluoride gravimetrically by precipitating as lead 
chlorofluoride? 

Barium. Strontium. Calcium. Magnesium 

(See also Probs. 278, 282, 285, 286, 295, 312, 315, 317, 318, 319, 323, 362, 
528, 537, 540, 545, 645, 667, 715, 717, 733, 891.) 

882. A sample of calcite (impure CaCO 3 ) weighing 1.402 grams is titrated 
with HC1 and requires 25.02 ml. What is the alkaline strength of the sample 
in terms of per cent CaO if 20.00 ml. of the HOI will just neutralize the NH 3 
that can be liberated from four millimoles of (NH 4 ) 3 PO4.2H 2 O? 

883. Given the following data: 35.27 ml. I 2 solution =c= 0.02991 gram As 2 O 3 ; 
30.00 ml. I 2 solution =0= 45.03 ml. Na 2 S 2 O 3 solution; 25.82 ml. Na 2 S 2 O 3 will 
reduce the iodine liberated from an excess of KI by 31.05 ml. KMnO 4 solution; 
15.42 ml. KMnO 4 <* 16.97 ml. KHC 2 O 4 .H 2 C 2 O 4 .2H 2 O solution; 1.000 nil. 
KIIC 2 O 4 .H 2 C 2 O 4 .2H 2 O solution =0= 1.074 ml. NaOH solution; 10.00 ml. NaOH 
solution =0 12.00 ml. HC1 solution. How many grams of CaCO 3 will be reacted 
upon by 29.83 ml. of this HC1 solution? 

884. A sample of Epsom salts is supposedly C.P. MgS0 4 .7H 2 O. On analysis 
of a sample weighing 0.8000 gram, the magnesium precipitated as MgNH 4 PO 4 
and ignited to Mg 2 P 2 O 7 was found to weigh 0.3900 gram. 

The sulfate precipitated as BaSO 4 weighed 0.8179 gram. 

(a) Is the sample chemically pure? (6) If not, is the sample contaminated 
with excess magnesium salt, excess sulfate, or excess water? (c) Is the 
magnesium equivalent to the sulfate? 

885. Basic magnesium carbonate corresponds approximately to the formula 
4MgCO 3 .Mg(OH) 2 .6H 2 O (F.W. = 503.7). The substance is sometimes 
roughly analyzed by determining its loss 011 ignition, but more generally by 
titration. 

A sample weighing 1.000 gram is dissolved in a 25-ml. pipetful of 1.000 N 
HC1 and the excess acid requires 5.01 ml. of 1.010 N NaOH. Calculate the 
percentage purity of the sample in terms of the above theoretical formula. 
What would be the loss on ignition of a 1.000-gram sample of the pure 
substance? 

886. A sample of dolomite is analyzed for Ca by precipitating as the oxalate 
and igniting the precipitate. The ignited product is assumed to be CaO, and 
the analyst reports 29.50 per cent Ca in the sample. Owing to insufficient 
ignition, the product actually contains 8.00 per cent of its weight of CaCO 3 * 
What is the correct percentage of Ca in the sample, and what is the per- 
centage error? 

887. A sample of magnesia limestone has the following composition: 

Silica = 3.00 per cent 

Ferric oxide and alumina = 0.20 per cent 
Calcium oxide = 33.10 per cent 

Magnesium oxide = 20.70 per cent 

Carbon dioxide = 43.00 per cent 



PROBLEMS 317 

In the manufacture of lime from the above, the carbon dioxide is reduced 
to 3.00 per cent. How many milliliters of 0.2500 N KMnO 4 will be required 
to determine the calcium volumetrically in a 1.000-gram sample of the lime? 

888. A sample of limestone containing 34.75 per cent Ca is given to a 
student for analysis. Using a 1.000-gram sample the student reports 35.26 
per cent Ca. If the error was due to insufficient ignition of the calcium 
oxalate precipitate causing contamination of the CaO by CaCO 3 , what was 
the percentage of CaCO 3 in the ignited product? What was the percentage 
error? What volume of sulfuric acid (sp. gr. 1.06, containing 8.77 per cent 
H 2 SO 4 by weight) should be added to this product to convert all the Ca into 
CaSO 4 ? What would be the new weight of the ignited product? 

Limestone. Lime. Cement 
(See also Probs. 246, 363, 364, 376, 378, 1025, 1026.) 

889. From the following data, compute the percentage of SiO 2 , A1 2 O 3> 
MgO, and CaO in a sample of cement weighing 0.6005 gram: Weight of 
SiO 2 = 0.1380 gram. Weight of Fe 2 O 3 + A1 2 O 3 = 0.1201 gram. Weight of 
Mg 2 P 2 O 7 = 0.0540 gram. Volume of 0.1429 N KMnO 4 for the Fe in the above 
ignited precipitate = 2.05 ml. Volume of this KMn0 4 required to titrate the 
precipitated calcium oxalate = 45.12 ml. 

890. A limestone contains only SiO 2 , CaCO 3 , FeC0 3 , MgCO 3 , MnCO 3 . 
Calculate the percentage of CO 2 from the following data: Sample = 0.800 
gram. Fe 2 3 + Mn 3 O 4 = 0.0521 gram and requires 5.00 ml. of 0.1112 N 
KMnO 4 for the iron. CaSO 4 = 0.7250 gram. Mg 2 P 2 O 7 = 0.0221 gram. 

891. A sample of limestone contains only silica, ferrous carbonate, calcium 
carbonate, and magnesium carbonate. From a sample weighing 1.200 grams 
there were obtained 0.0400 gram of ignited ferric oxide, 0.5003 gram of CO 2 , 
and 0.5007 gram of magnesium pyrophosphate. Find the volume of ammonium 
oxalate solution [containing 35.00 grams of (NH 4 ) 2 C 2 O 4 .H 2 O per liter] required 
to precipitate the calcium as oxalate. Also calculate the normality of KMnO 4 
if 38.00 ml. are required to titrate the oxalate precipitate. 

Iron. Aluminum. Titanium 

(See also Probs. 245, 251, 271, 281, 284, 289, 293, 294, 301, 302, 303, 304, 
305, 309, 320, 366, 375, 639, 641, 646, 651, 657, 668, 672, 673, 675, 677, 703, 
764, 810.) 

892. How many milliliters of ammonium hydroxide (sp. gr. 0.946, con- 
taining 13.88 per cent NH 3 by weight) are required to precipitate the iron as 
Fe(OH) 3 from a sample of pure FeSO 4 .(NH 4 ) 2 SO 4 .6H 2 O which requires 0.34 ml. 
of hot HNO 3 (sp. gr. 1 .350, containing 55.79 per cent HNO 3 by weight) for 
oxidation? Assume reduction of HNO 3 to NO. 

893. A sample of magnetite (impure Fe 3 4 ) is fused with an oxidizing flux, 
leached with water, and acidified. The solution is divided into two equal 
portions. 

In one portion the iron is precipitated as Fe(OH) 3 and ignited in the regular 
way. What was the weight of the original sample if the number of centigrams 



318 CALCULATIONS OF ANALYTICAL CHEMISTRY 

of ignited precipitate is found to be one-third the percentage of FesO 4 in the 
sample? 

In the other portion the iron is reduced with zinc and titrated with KMnO 4 . 
What should be the normality of the KMnO 4 so that the percentage of Fe 8 O 4 
will be twice the buret reading? 

894. A 1.000-gram sample of limonite containing inactive impurities is dis- 
solved in acid, and the solution is divided into two equal portions. One 
portion is reduced and titrated with KMnO 4 (1.000 ml. =0= 0.008193 gram 
H 2 C 2 O4.2H 2 O). The other portion is just neutralized, and 40.00 ml. of 1.500 N 
ammonia are added to precipitate the iron. This is in excess of the necessary 
amount, and the number of milliliters in excess is equal to the number of 
milliliters of KMnO 4 required in the volumetric process. What is the per- 
centage of Fe in the sample? 

896. If 60.00 ml. of BaCl 2 (0.1000 N as a precipitating agent) are required 
just to precipitate all the sulfate from a sample of pure ferric alum, Fe 2 (SO 4 ) 3 . 
(NH 4 ) 2 SO 4 .24H 2 O, how many milliliters of NH 4 OH (sp. gr. 0.900, containing 
28.33 per cent NH 3 by weight) would be required just to precipitate all the 
iron from the same weight of sample? 

896. What is the percentage purity of a sample of ferrous sulfate FeSO 4 .7H 2 O 
weighing 1.000 gram, if, after it has been dissolved in water, 10.00 ml. of 
0.1100 N hydrochloric acid have been added to it, and it has been oxidized 
with bromine, 11.73 ml. of N NH 4 OH are required to neutralize the acid and 
precipitate the iron as Fe(OH) 3 ? 

897. How many milliliters of 0.1250 N KMnO 4 are needed to titrate a solu- 
tion containing ferrous iron if by a gravimetric method 3.50 ml. of 6.00 N 
ammonia water are required to precipitate the iron after oxidation to the ferric 
condition? 

898. What weight of mineral containing ferrous iron should be taken for 
analysis so that twice the number of milliliters of permanganate used for 
oxidation (20.0 ml. o 30.0 ml. of potassium tetroxalate solution which is 
0.0400 N as an acid) will be three times the percentage FeO? With this 
weight of sample, 15.0 ml. of the permanganate are required, and the water 
evolved by strong ignition weighs 0.0256 gram. What is the percentage 
of FeO in a finely ground sample in which the percentage of water is 1.05? 

899. After decomposition of a half-gram sample of a certain mineral and 
the removal of silica, the addition of bromine and NH 4 OH precipitates 
Fe(OH) 3 + A1(OH) 3 . On ignition, these weigh 0.1205 gram. They are then 
fused with KHS0 4 , dissolved in dilute H 2 SO 4 , passed through a column of 
amalgamated zinc, and the iron titrated with KMnO 4 (1.000 ml. =0= 0.02500 
millimole Na 2 C 2 O 4 ), requiring 22.46 ml. What is the percentage of A1 2 O 3 
and of FeO in the original sample? How many milliliters of NH 4 OH (sp. 
gr. 0.970, containing 7.31 per cent NH 3 by weight) were required just to 
precipitate all of the ferric iron and aluminum from the solution after neu- 
tralizing the acid? 



PROBLEMS 319 

900. " Iron by hydrogen " is obtained by reducing pure Fe 2 Os with hydrogen. 
It is a fine gray powder used analytically as a reagent for the determination 
of nitrates by reduction to ammonia. The material should contain at least 
90 per cent metallic iron and is generally contaminated with an oxide assumed 
to be Fe 3 O 4 , as the reduction is not complete. The metallic iron in the sample 
is soluble in a neutral solution of FeCl 3 , according to the equation Fe + 
2Fe+++ > 3Fe++, and the ferrous chloride formed is determined by titration 
with KMnO 4 . 

In an actual analysis, 0.5000 gram is weighed into a 100-ml. measuring 
flask, the air displaced with C0 2 , and water added, 2.500 grams (an excess) 
of anhydrous FeCls (free from Fe++) are added, and the flask stoppered and 
shaken for 15 minutes. The solution is diluted to the mark, mixed, and 
filtered. Twenty milliliters of the filtrate are titrated with 44.16 ml. of 
0.1094 N KMnO 4 after the addition of sulfuric acid and manganese sulfate 
titrating solution and proper dilution. Calculate the percentage of metallic 
iron in the sample. 

901. A sample of aluminum sulfate is known to be contaminated with 
iron and manganese. A sample weighing 3.362 gram is dissolved in dilute 
acid, and bromine and ammonia are added to precipitate A1(OH) 3 , Fe(OII) 3 
and MnO 2 . Treating the precipitate with concentrated HNO 8 dissolves 
the iron and aluminum hydroxides and leaves the MnO 2 . This MnO 2 is 
ignited in air (forming Mn 3 O4) and the product is found to weigh 0.0363 gram. 
The HNOs solution is evaporated with H 2 SO 4 and the iron eventually reduced 
and titrated with 4.90 ml. of 0.1020 N KMnO 4 . An acid solution of 3.829 
grams of the original salt gives with bromine and ammonia a precipitate that 
on ignition in air weighs 0.5792 gram. What is the percentage of Al 2 Oa 
and of Mn and Fe in the original material? 

902. A volumetric method for aluminum has been found useful in certain 
cases. The aluminum is precipitated with 8-hydroxyquinoline ("oxine") 
and the precipitate is dissolved in acid and titrated with standard KBrOs 
(+KBr) (See Part VI, under Aluminum). If an excess of KBr is used and the 
titration requires 48.0 ml. of KBrO 3 (of which 1.00 ml. will liberate from excess 
KI in the presence of acid sufficient I 2 to require 1.00 ml. of 0.100 N Na 2 S 2 O 3 ), 
what weight of residue would have been obtained if the oxine precipitate had 
been ignited in air? 

903. A solution of ferric chloride is prepared by dissolving 10.03 grams of 
pure iron in HC1, oxidizing, and diluting to a liter. If 50.0 ml. of this solu- 
tion react with 40.0 ml. of a TiCl 3 solution, what is the normality of the latter 
as a reducing agent? 

904. A sample of titanium ore is treated in such a way that all the iron is 
present in the 2-valent condition and all the titanium in the 3-valent condition. 
The solution is then titrated with ferric alum solution of which 50.00 ml. yield 
0.4000 gram of Fe 2 O 3 . If the original sample weighed 0.6000 gram and 15.00 
ml. of ferric alum solution were used, find the percentage of TiO 2 in the ore. 

906. The iron in a solution of a 0.800-gram sample of titanium ore was 



320 CALCULATIONS OF ANALYTICAL CHEMISTRY 

reduced with stannous chloride and then reacted with 26.0 ml. of KMnC>4 
(1.00 ml. =0= 0.800 ml. of potassium tetroxalate solution which is 0.08000 N 
as an acid). The sulfuric acid solution of the same weight of sample was re- 
duced with zinc, and the reduced solution was caught in an acid solution of 
ferric alum which then reacted with 48.0 ml. of the above KMnC>4. Compute 
the analysis on the basis that the original sample contained only Fe 3 O 4 , TiO 2 , 
and SiO 2 . (Zinc reduces Ti from valence 4 to 3; SnCl 2 does not.) 

906. A silicate rock is shown to contain ferrous iron, aluminum, and 
titanium. A sample weighing 0.6050 gram is decomposed by an oxidizing 
flux, the silica removed, and the precipitate obtained by NH 4 OH filtered 
off, this precipitate consisting of the hydroxides of ferric iron, aluminum, 
and titanium. The ignited precipitate weighs 0.5120 gram. It is fused with 
K 2 S 2 O7, and brought into solution, and the solution divided into two equal 
portions. One portion is poured through a Jones reductor containing amal- 
gamated zinc and the solution caught directly in excess ferric alum solution. 
This solution is then titrated with 0.08333 N KMnO 4 , of which 19.56 ml. are 
required. The other portion is reduced with SnCl 2 , the excess stannous is 
destroyed, and the solution titrated with 0.08333 N KMnO 4 , of which 1 1.94 ml. 
are required. Calculate the percentages of FeO, A1 2 O 3 , and TiO 2 in the original 
silicate. 

Cerium. Thorium. Zirconium. Uranium. Beryllium. Bismuth. Boron 
(See also Probs. 273, 371.) 

907. From the methods given for the determination of cerium in Part VI, 
outline a possible iodimetric method for that element and indicate the correct 
milliequivalent weight to be used. 

908. A solution of uranyl nitrate is divided into two equal parts. One 
portion is evaporated to fumes with H 2 SO 4 , diluted, and passed through a 
Jones reductor. The uranium forms uranous ions (U ++ " H ~). The solution is 
titrated with 0.1200 N KMnO 4 , requiring 20.50 ml. The uranium in the other 
portion is precipitated with NIT 4 OH as (NH 4 ) 2 U 2 O7 and the precipitate is 
ignited and weighed (see Part VI). Write equation for the titration and for 
the ignition (NH 3 and N 2 are among the products in the latter case) and cal- 
culate the weight of residue obtained in the ignition. 

909. An iodimetric method for determining zirconium has been suggested. 
The element is precipitated with selenious acid (H 2 SeO 3 ) as ZrOSeO 3 . The 
precipitate is dissolved, treated with KI (SeOr + 4I~ -> Se + 2I 2 + 3H 2 O) 
and the liberated iodine subsequently titrated with standard Na 2 S 2 O 3 . If 
5.00 ml. of 0.0833 N Na 2 S 2 O 3 were required for a given weight of sample by 
this method, how many grams of residue would be obtained by igniting the 
ZrOSeO 3 obtained from another sample of the same weight (see Part VI, 
under Zirconium)? 

910. Anhydrous sodium tetraborate reacts with water according to the 
following equation: 

B 4 O 7 - + 5H 2 O - 2H 2 BO 3 - + 2H 3 BO 3 



PROBLEMS 321 

The anhydrous salt may be dissolved in water and titrated directly by means 
of hydrochloric acid, with methyl orange as indicator, as represented by the 
following equation: 

H 2 BO 3 - + H+ - H 3 BO 8 

After conversion of the sodium tetraborate to boric acid by careful titration 
with HC1, methyl orange indicator being employed, a suitable polyhydric 
alcohol (inverted sugar, glycerin, mannitol, etc.) is added and the first hy- 
drogen of the boric acid is titrated by means of standard sodium hydroxide, 
with phenolphthalein indicator. 

The following data were obtained on a sample of tetraborate: 

Sample = 0.3000 gram 

Volume of IIC1 required, with methyl orange indicator = 26.35 ml. 
Volume of NaOH required to titrate the boric acid = 58.10 ml. 
Normality NaOH = 0.1030 
40.00 ml. NaOH o 36.35 ml. HC1 

Calculate the percentage of Na 2 B 4 O7 present in the sample (a) using the 
data from the acid titration, (6) using the data from the alkali titration. 

911. Barium can be determined volumetrically (after precipitating as 
BaCrO 4 ) either by a permanganate process or by an iodi metric process (see 
Part VI, under Barium). If in the permanganate process 25.00 ml. of 0.1000 N 
ferrous ammonium sulfate were used and the excess ferrous required 10.50 ml. 
of 0.06667 N KMnO 4 , what weight of BaSO 4 would be precipitated during the 
titration reactions? If the iodimetric method had been used on the same 
weight of sample, how many milliliters of 0.06667 N Na 2 S2O3 would have been 
required? 

Ans. 0.1400 gram. 27.00 ml. 

912. Beryllium can be determined volumetrically by precipitating with 
oxine and titrating with KBrO 3 as in the case of aluminum (see Part VI, under 
Aluminum). Write the corresponding equations for the behavior of beryllium 
and calculate the weight of the ignited residue (of BeO) using the same 
numerical data as given in Prob. 902. 

913. If bismuth is determined by precipitating as bismuthyl oxalate and 
titrating with standard KMnO 4 (see Part VI, under Bismuth), what is the 
value of each milliliter of 0.1000 N KMnO 4 in terms of grams of Bi 2 O 3 ? 

914. In the volumetric method for boron (see Part VI), what is the milli- 
equivalent weight of B 2 O 3 ? Look up the ionization constant for boric acid 
and plot the titration curve to show its general appearance. Show how it 
compares in appearance with the curve for the titration of HC1 under similar 
conditions of concentration. 

Copper. Lead. Zinc. Cadmium. Brass 

(See also Probs. 248, 261, 264, 274, 279, 332, 336, 337, 339, 342, 344, 345, 
347, 349, 350, 352, 355, 356, 357, 358, 359, 360, 522, 640, 681, 697, 704, 711, 
728, 739, 751, 1005, 1006, 1007, 1010, 1017.) 



322 CALCULATIONS OF ANALYTICAL CHEMISTRY 

916. How many milliliters of HNO 8 (sp. gr. 1.130, containing 21.77 per cent 
HNO 3 by weight) are theoretically required to dissolve 5.00 grams of brass 
containing 0.61 per cent Pb, 24.39 per cent Zn, and 75.00 per cent Cu? Assume 
reduction of the HNO 3 to NO by each constituent. What fraction of this 
volume of acid is used for oxidation? 

916. What volume of sulfuric acid (sp. gr. 1.420) is required to displace 
the nitrate radical from the mixture of salts obtained by dissolving 25.00 grams 
of brass (68.29 per cent Cu, 31.50 per cent Zn, 0.21 per cent Pb) in nitric acid 
and evaporating to dryness? 

917. What is the percentage of copper in a steel if with a 5.00-gram sample 
the volume of H 2 S gas (measured under standard conditions) required to 
precipitate the copper as CuS is 2.00 ml. more than the volume of 0.100 N 
Na 2 S 2 O 3 solution subsequently required for the copper by the iodimetric 
method? 

918. If 0.800 gram of a lead ore yields a precipitate of chromate that 
contains chromium sufficient to yield on treatment with an excess of KI in 
acid solution an amount of iodine to react with 48.0 ml. of 0.1000 N thiosulfate 
solution, find the percentage of Pb in the ore. 

919. What weight of zinc ore should be taken for analysis such that the 
number of milliliters of 0.1000 molar ferrocyanide solution used will equal the 
percentage of Zn in the ore? 

920. If a copper ore on being analyzed yields 0.235 gram of Cu 2 S after 
being heated with sulfur in a stream of hydrogen, how many grams of KIO 3 
would react in the iodate method with the same weight of ore? 

92L If 0.5000 gram of a copper alloy containing 25.00 per cent Cu requires 
20.00 ml. of KCN for titration, what is the equivalent of 1.000 ml. of the KCN 
(a) in terms of Ag (using KI as indicator) and (b) in terms of Ni? How many 
milliliters of KIO 3 solution would have been required by the iodate method 
if with an excess of KI, 15.00 ml. of the KIO 3 would have liberated I 2 enough to 
react with a volume of 0.1000 N thiosulfate equivalent to 0.1000 gram of 
K 2 Cr 2 O 7 ? 

922. A brass weighing 0.800 gram contains 75.02 per cent Cu, 23.03 per cent 
Zn, and 1.95 per cent Pb. What volume of 0.1000 N Na 2 S 2 O 3 would be used in 
the determination of copper by adding KI and titrating the liberated iodine? 
What volume of 0.05000 N KMnO 4 would be required for the lead if it is pre- 
cipitated as chromate, dissolved, reduced with 25.00 ml. of 0.04000 N FeSO 4 , 
and the excess ferrous ions titrated with the KMnO 4 ? What weight of zinc 
pyrophosphate would be obtained in the determination of zinc? 

923. In a certain volumetric method for determining copper, the element is 
precipitated as CuCNS and the precipitate is titrated with standard KIO 8 
according to the net equation: 4CuCNS + 7KIO 3 + 14HC1 - 4CuS0 4 + 
7IC1 + 4HCN + 7KC1 + 5H 2 0. If the KIO 3 is of such concentration that 
1.000 ml. will liberate from excess KI in the presence of acid sufficient iodine 



PROBLEMS 323 



to react with 1.000 ml. of 0.1000 N Na^Oa, what is the value of 1.000 ml. of 
the KIOs in terms of grams of Cu in the above method? 
Ans. 0.0006054 gram. 

924. If a solution contains that amount of Cu++ requiring 10.0 ml. of a 
Naj&Os solution in the common iodimetric method for copper, how many 
milliliters of KIO 3 solution (1.00 ml. =c= 2.00 ml. of the above Na^Os) would 
be required to reach that point in the titration by the iodate method (see 
Part VI, under Copper) corresponding to the maximum intensity of color of 
the CHC1 3 indicator? 

Ans. 42.0 ml. 

925. Red lead (Pb 3 O 4 ) is made by the direct oxidation of metallic lead. 
Chemically it may be regarded as 2PbO.PbO 2 , but owing to uneven heating 
in the manufacturing process commercial samples vary somewhat from the 
theoretical composition and are likely to contain excess Pb0 2 . When such 
samples are treated with dilute HNO 3 , the monoxide dissolves, leaving a 
brown residue H 2 PbO 3 , which can be reduced by the addition of a measured 
excess of oxalate ion and the excess titrated with standard permanganate. 

A sample weighing 0.7000 gram is treated with dilute HNO 3 and subse- 
quently with a weighed amount of pure sodium oxalate (5.000 milliequivalents). 
After complete reaction, the solution is diluted with boiling water, manga- 
nous sulfate solution added, and the excess oxalate ion titrated with 28.56 ml. 
of 0.09987 N permanganate. Calculate: (a) the total percentage of PbOa 
(free and combined) in the sample; (6) the oxidizing power to percentage of 
Pb 3 O 4 ; (r) on the assumption that the sample is composed only of 2PbO.PbC>2 
and excess Pb0 2 , the percentage of each. 

Ans. (a) 36,70 per cent total PbO 2 ; (b) 105.2 per cent Pb 3 O 4 ; (c) 97.18 
per cent Pb 3 (X, 2.82 per cent free PbO 2 . 

926. A solution of (NH 4 ) 2 IIPO 4 is made up to be 2.00 N as an ammonium 
salt. Calculate'approximately the volume necessary to precipitate the zinc as 
ZnNH 4 PO4 in a sample of brass (90.0 per cent Cu, 10.0 per cent Zn) weighing 
5.00 grams. 

927. If in the analysis of a brass containing 28.0 per cent Zn an error is 
made in weighing a 2.500-gram portion by which 0.001 gram too much is 
weighed out, what percentage error in the zinc determination would be 
made? What volume of a solution of diammonium phosphate, containing 
90.0 grams of (NH 4 ) 2 HPO 4 per liter, would be required to precipitate the zinc 
as ZnNH 4 PO 4 , and what weight of precipitate would be obtained? 

928. In the electrolysis of a sample of brass weighing 0.8000 gram, there 
are obtained 0.0030 gram of PbO 2 , and a deposit of copper exactly equal in 
weight to the ignited precipitate of Zn 2 P 2 O 7 subsequently obtained from the 
solution. What is the percentage composition of the brass? 

Tin. Antimony. Arsenic. Bronze 

(See also Probs. 263, 270, 287, 297, 310, 321, 348, 692, 694, 698, 699, 708, 
709, 719). 



324 CALCULATIONS OF ANALYTICAL CHEMISTRY 

929. A sample of stibnite weighs 0.5000 gram. The percentage of Sb as 
found by titration in neutral solution with 0.1000 N iodine was 30.00 per cent. 
If the buret reading as recorded was 0.45 ml. too large, what was the true 
percentage and what was the percentage error? 

930. A sample of type metal weighing 1.100 grains is dissolved in con- 
centrated H 2 SO4. Concentrated HC1 is added to the cooled solution, and the 
solution is boiled. At this point antimony is in the 3-valent state; tin is in the 
4-valent state. The antimony in the cold solution is then titrated rapidly with 
KMn0 4 (1 ml. =c= 0.00558 gram Fe), requiring 32.80 ml. More HC1 is added, 
and the solution is boiled with powdered lead which reduces the antimony from 
valence 5 to valence 3 and the tin from valence 4 to valence 2. The tin is then 
quickly titrated in cold acid solution with I 2 (1 ml. =0= 0.0500 millimole As 2 O 3 ), 
requiring 9.27 ml. What are the percentages of Sb and Sn in the alloy? If 
the same weight of alloy had been treated with 6 N HNO 3 and the residual 
metastannic and antimonic acids had been ignited, what weight of product 
would have been obtained? 

931. A sample weighing 0.250 gram and containing arsenic is dissolved, and 
the solution containing the trivalent element is electrolyzed (method of 
Hefti). Arsine is liberated and is conducted into 50.0 ml. of 0.125 N iodine 
solution. The excess of the latter reacts with 20.0 ml. of Na 2 S 2 O 3 solution, of 
which 1.00 ml. = 0.00500 gram of copper. Find (a) percentage of As 2 O 3 in 
the sample and (b) the time required for electrolysis if a current of 3.00 amperes 
is used and only 40.0 per cent of the current is used in reducing the arsenic. 

932. An alloy containing arsenic weighs 5.10 grams. The arsenic is distilled 
as AsCI 3 from a strong HC1 solution of the alloy and eventually titrated in 
nearly neutral bicarbonate solution with standard iodine (1.00 ml. =0= 1.00 ml. 
Na 2 S2Oa == 0.0024 gram Cu). Calculate the percentage of arsenic in the alloy 
if 5.00 ml. are required. If the arsenic were evolved as arsine and the arsirie 
absorbed in excess 0.100 N iodine (which oxidizes the arsenic to arsenate), 
how many milliliters of 0.0833 N Na 2 S 2 O 3 would be equivalent to the iodine 
used up? 

933. A mixture of As 2 3 + As 2 O 5 + inert matter is dissolved and titrated 
in neutral solution with I 2 [1.00 ml. =0= 1.00 ml. KMnO 4 =0= 0.0500 millimole 
FeSO4.(NH 4 ) 2 SO4.6H 2 O] requiring 20.00 ml. The resulting solution is acidified 
and an excess of KI is added. The liberated I 2 requires 30.50 ml. of Na 2 S 2 O 3 
[1.00 ml. o 0.0100 millimole of KH(IO 3 ) 2 ]. Calculate the weight in grams of 
combined As 2 C>3 + As 2 6 in the sample. 

Carbon. Carbon Dioxide. Silicon. Tungsten. Molybdenum 
(See also Probs. 272, 276, 277, 515, 794, 796, 797, 798, 804, 805, 808, 890.) 

934. If a 1.30-gram sample of iron containing 1.15 per cent carbon is 
analyzed by direct combustion, what would be the gain in weight of the ab- 
sorption tube? If the gas is passed through 100 ml. of 0.0833 N Ba(OH) 2 , 
how many milliliters of 0.100 N HC1 would be required to titrate the super- 
natant liquid? 



PROBLEMS 325 

936. A sample of steel weighing 1.00 gram is burned in oxygen. The 
CO2 is caught in a 100-ml. pipetful of Ba(OH) 2 solution. The supernatant 
liquid requires 96.50 ml. of 0.100 N HC1. If the steel contains 0.57 per cent car- 
bon, what is the normality of the barium solution used and how many grams 
of Ba(OH) 2 .8H 2 O are contained in each milliliter? 

936. What is the percentage of carbon in a 5.00-gram sample of steel if on 
combustion the ascarite tube gains 0.1601 gram in weight? Using the same 
weight of sample and passing the gas into Ba(OH) 2 solution, what must be 
the normality of an HC1 solution so that the milliliters required to titrate the 
BaC0 3 precipitate will be twenty-five times the percentage of C? 

937. A 2-gram sample of steel is burned in oxygen, and the evolved CO 2 
after passing through appropriate purifying trains is caught in 100 ml. of 
Ba(OH) 2 solution. The supernatant liquid requires 75.0 ml. of HC1 [1.00 
ml. =0= 0.00626 gram Na 2 CO 3 ; 1.00 ml. o 1.12 ml. of the Ba(OH) 2 solution]. 
What is the percentage of carbon in the steel and what would have been the 
gain of an ascarite bulb if a similar sample had been analyzed by the absorption 
method? 

938. What volume of 6.00 N hydrofluoric acid is theoretically required to 
volatilize the silica from 0.5000 gram of KAlSi 3 O 8 ? What volume of SiF 4 
at 29C. and 765 mm. pressure is produced? 

939. A 3.00-gram sample of steel contains 3.00 per cent Fe 2 Si. After it has 
been dissolved in HNO 3 and evaporated, what weight of SiO 2 will be obtained? 
What volume of SiF4 under standard conditions will be evolved by the action 
of 1IF + H 2 SO 4 on the SiO 2 ? 

940. A 3.00-gram sample of steel containing 1.21 per cent Si and 0.23 per cent 
W is dissolved in concentrated 1INO 3 and evaporated to dryness. What should 
be the weight of the ignited acid-insoluble residue before and after treatment 
with HF? 

941. A sample of tungsten steel weighing 5.000 grams is] dissolved in aqua 
regia, evaporated to dry ness, and dehydrated. The acid-insoluble residue 
weighs 0.0928 gram and after treatment with HF weighs 20.00 per cent less. 
What are the percentages of Si and of W in the steel? 

942. On the assumption that Mo 24 O 3 7 is a mixture of MoO 3 and Mo 2 O 3 , 
what percentage of the total Mo is in the 3-valent state and what percentage 
is in the 6-valent state? 

943. What weight of molybdenum steel should be taken so that 1.00 ml. of 
0.0600 N KMnO 4 will be used for each 0.50 per cent of Mo, on the basis of 
reduction of element to the trivalent condition and reoxidation by the per- 
manganate to the oxidation number of 6? 

Chromium. Vanadium 
(See also Probs. 296, 654, 656, 666, 678, 775, 1012.) 

944. From the following data compute the percentage of Cr in a sample of 
steel. Weight of sample = 1.850 grams. After the chromium has been 



326 CALCULATIONS OF ANALYTICAL CHEMISTRY 

oxidized to dichromate with KMnO4 and the excess reagent removed, 150 ml. 
of 0.0800 N ferrous solution are added and the solution then reacts with 14.00 
ml. of 0.0900 N KMnO 4 . 

945. It is desired to prepare a solution of chromium acetate to contain 
8.00 per cent Cr 2 3 by weight for use as a mordant. A batch of the material 
is made up to the approximate concentration and is found to have a specific 
gravity of 1.195. A 2.000-ml. sample is taken, and the chromium is oxidized 
to dichromate. To one-half the solution are added 50.00 ml. of ferrous sulfate 
solution, and the excess ferrous iron requires exactly 17.32 ml. of 0.1334 N 
KMnO 4 for oxidation (25.00 ml. FeSO 4 solution =c= 21.73 ml. KMnO 4 solution). 
How many pounds of water must be evaporated from one ton of the liquor 
to give the desired concentration? 

946. A steel containing 0.90 per cent Cr weighs 2.000 grams. The chromium 
is oxidized to chromate, and to the acidified solution is added an excess of KI. 
The liberated iodine requires 10.00 ml. of thiosulfate solution. What is the 
normality of the thiosulfate solution? 

947. Assuming that vanadium like nitrogen forms five oxides and that any 
other oxide is a mixture of two or more of these, compute the oxidation number 
of the vanadium in the reduced condition and show what combination of oxides 
could give this. Use the following data: 0.08500 gram of Na 2 V 4 O 9 after an 
abnormal reduction was oxidized to the 5-valent condition by 43.14 ml. of 
KMnO 4 of which 40.00 ml. reacted with 30.00 rnl. of potassium tetroxalate 
solution which was 0.08000 N as an acid. 

948. If 0.394 gram of Na 2 V 4 O 9 is reduced and requires 10.00 ml. of per- 
manganate (1.000 ml. o 0.0536 gram Na 2 C 2 O 4 ) to oxidize the vanadium back 
to vanadic acid, find the valence of the reduced vanadium. 

949. A sample of chrome-vanadium steel weighing 2.00 grams is dissolved 
in H 2 SO 4 + H 3 PO 4 , and HNO 3 is added to oxidize the iron and carbides. 
In the presence of silver ions (catalyst), ammonium persulfate is added to 
oxidize chromic ions to dichromate, vanadyl ions to metavanadate, and man- 
ganous ions to permanganate. Excess persulfate is destroyed by boiling, 
and the permanganate is reduced with a small quantity of HC1. The addition 
of 25.0 ml. of 0.1010 N FeSO 4 causes reduction of vanadate and dichromate, 
and the excess ferrous and the reduced vanadium are titrated with 0.1120 N 
KMnO 4 , of which 12.6 ml. are required. A small amount of FeSO 4 is added to 
reduce the vanadium again and the excess ferrous ions destroyed with per- 
sulfate. The vanadium alone is then titrated with the above-mentioned 
KMnO 4 , of which 0.86 ml. are required. Write ionic equations for all chemical 
changes involving the above-mentioned elements, and calculate the percentage 
of Cr and of V in the sample. 

960. The determination of vanadic acid (HV0 3 ) in the presence of molybdic 
acid (H 2 MoO 4 ) depends upon the fact that vanadic acid alone is reduced to 
VO++ by sulfur dioxide in dilute sulfuric acid and can be reoxidized by standard 
permanganate solution. Both vanadic acid and molybdic acid are reduced by 
amalgamated zinc, the former to V* 4 " and the latter to Mo+ ++ . These reactions 



PROBLEMS 327 

are carried out in a Jones redactor and the reduced constituents oxidized by 
being passed into an excess of ferric salt and phosphoric acid. An equivalent 
reduction of the ferric iron to ferrous takes place. The ferrous iron is then 
titrated with standard permanganate. 

REDUCTION WITH SO2 REDUCTION WITH Zn 

Grams of sample = 0.4500 Grams of sample = 0.4500 

Normality of KMnO 4 = 0.1092 Normality of KMnO 4 = 0.1092 

Ml. KMnO 4 = 8.23 Ml. KMnO 4 = 41.74 

a. Complete and balance all the equations in this process. 
6. Express the amount of vanadate as percentage of V and the amount 
of molybdate as percentage of Mo. 

951. Calculate the percentage of chromium and of vanadium in a chrome- 
vanadium steel from the following data: 

Chromium. A sample weighing 2.00 grams is dissolved in an acid mixture. 
Subsequent treatments convert iron to Fe +4 " + , manganese to Mn+ +, chromium 
to Cr 2 O7~ and vanadium to VO 3 ". A 25-ml. pipetful of standard ferrous solu- 
tion [39.2 grams of FeSO4.(NII 4 ) 2 SO4.6H 2 O per liter] is added. Chromium is 
thus reduced to Cr+++, vanadium to VO++. The solution is titrated with 
KMnO 4 (1.00 ml. =0= 0.92 ml. of the ferrous solution) requiring 14.28 ml. to 
give a permanent pink color. Only vanadium and the excess ferrous iron are 
oxidized in this step. To correct for overtitration and color interferences, the 
solution is boiled until the permanganate color is destroyed, and the solution 
is brought to the same shade of color as before with the standard perman- 
ganate, requiring 0.08 ml. 

Vanadium. The vanadium in the above solution is now reduced to VO 4 " 4 
with dilute FeSO 4 solution and the excess ferrous oxidized with a small amount 
of persulfate. The vanadium is then titrated back to VO 3 ~ with the above 
KMnO 4 , requiring 1.10 ml. It may be assumed that the solution is overtitrated 
to the same degree as in the first titration. 

Manganese 

(See also Probs. 266, 269, 351, 370, 379, 642, 643, 648, 652, 653, 655, 664, 
671, 682, 691, 701.) 

962. What weight of pyrolusite containing 75.0 per cent MnO 2 will oxidize 
the same amount of oxalic acid as 35.0 ml. of a KMnO 4 solution of which 
1. 00 ml. will liberate 0.0175 gram of iodine from an excess of potassium 
iodide? 

953. An oxide of Mn weighing 0.4580 gram is treated with dilute H 2 SO 4 
and 50.00 ml. of 0.1000 N ferrous ammonium sulfate solution. After the 
reduction of the manganese to the manganous condition is complete, the excess 
of ferrous solution reacts with 30.00 ml. of 0.03333 N KMn0 4 . Find the 
symbol of the original oxide of Mn. 

964. Given the following data in the analysis of pyrolusite by the iodimetric 
process, find the volume of disodium phosphate solution (90.0 grams of 



328 CALCULATIONS OF ANALYTICAL CHEMISTRY 

Na 2 HPO 4 .12H 2 O per liter) that would be necessary to precipitate the Mn as 
MnNH 4 P0 4 from 0.5000 gram of the sample. 

Weight of sample = 1 .000 gram 
Na 2 S 2 O 3 solution = 40.40 ml. 

The thiosulfate solution is equivalent in reducing power to a stannous 
chloride solution that contains 29.75 grams of tin per liter. 

956. What volume of bromine water (30.0 grams Br 2 per liter) would 
theoretically be required to precipitate the manganese from an acetic acid 
solution of its salt, if the resulting precipitate of Mn0 2 gives on ignition 
0.1060 gram of Mn 3 O4? How many milliliters of sulfurous acid (sp. gr. 
1.028, containing 5.00 per cent SO 2 by weight) would be required to dissolve 
the MnO 2 precipitate, and what weight of Mn 2 P 2 O 7 would be obtained from 
the resulting solution? 

966. If the manganese in 50.0 ml. of 0.0833 N KMnO 4 solution were reduced 
to the manganous condition, how many milliliters of 0.0833 N KMnO 4 would 
be equivalent to the Mn in the reduced solution by the (a) Volhard method, 
(b) bismuthate method, and (c) chlorate method? 

967. It has been shown that manganous ions can be titrated potentio- 
metrically with standard KMiiO 4 in nearly neutral pyrophosphate solution 
according to the equation: 4Mn++ + MnO 4 - + 8H+ + 15II 2 P 2 O 7 " - 
5Mn(H 2 P2O 7 )s- + 4H 2 O. What is the value of each millilitcr of KMnO 4 in 
terms of Mn by this method if each milHliter of the KMnO 4 is equivalent to 
0.002040 gram of sodium formate (NaCIIO 2 ) when titrated according to the 
equation: 3CHO," + 2Mn(V + H 2 -> 2MnO 2 + 3CO 2 + 5OH~? 

Ans. 0.004394 gram. 

968. When a constituent is to be determined with extreme accuracy, 
e.g., in a case where the manganese content of a sample of steel may be in 
dispute, it is best to analyze for the constituent by two different methods. 
It is also best to standardize the solutions used against a sample of like material 
of known composition rather than in the usual way. This procedure gives a 
direct comparison under identical conditions with a standard. 

The purchaser of a quantity of steel has reserved the right to reject the lot, 
which is to be used for a special purpose, if the manganese content is less than 
0.350 per cent. The sample must therefore be prepared and the analysis made 
with extreme accuracy. On the basis of a representative sample, the method 
and results of an analysis follow. Calculate to three significant figures (a) the 
percentage of manganese in the steel by both methods, (6) the normality of the 
permanganate. 

Persulfate Process. A 0.1000-gram sample of Bureau of Standards steel 
containing 0.660 per cent Mn required 7.03 ml. of arsenite solution. A 
similar-weight sample of the steel under investigation required 4.36 ml. 

Bismuthate Process. A one-gram sample of the above-mentioned Bureau 
of Standards steel was used. A pipetful of FeSO 4 was used, and the excess 
ferrous salt required 16.96 ml. of standard KMnO 4 . A one-gram sample 



PROBLEMS 329 

of the steel under investigation, treated the same way, required 19.63 ml. 
of the KMnO 4 . One pipetful of FeSO 4 ~ 23.97 ml. of KMnO 4 . 

959. A carefully prepared steel is to be used as a standard in subsequent 
analyses of other steels for manganese by the persulfate method. To determine 
the correct percentage of manganese in the standard steel a sample weighing 
1.05 grams is analyzed by the bismuthate method. A 25-ml. pipetful of ferrous 
ammonium sulfate is used, and the titration requires 13.2 ml. KMnO 4 (1 ml. =c= 
0.00201 gram Nu 2 C 2 O 4 ; 1 ml. =c= 1.02 ml. of the ferrous solution). What is the 
percentage of manganese in the steel? 

In the routine analysis of a certain plain carbon steel by the persulfate 
method, the analysis is run in parallel with a corresponding analysis of the 
above standard steel. The same weights of sample are used in the two cases. 
The above standard steel requires 10.4 ml. of arseiiite solution; the unknown 
steel requires 17.1 ml. What is the percentage of manganese in the latter 
steel? If in the persulfate method a one-gram sample was used and the arsenite 
solution contained 1.10 grams of As 2 O 3 per liter, to what average oxidation 
number was the manganese reduced in the titration? 

A/is. 0.355 per cent. 0.584 per cent. 3.4. 

Cobalt. Nickel 
(See also Prob. 334, 742, 743, 749, 750.) 

960. A sample of ore weighing 0.8900 gram yields by electrolysis 0.2670 
gram of Ni and Co, and from the deposited metal a precipitate weighing 
0.9405 gram is obtained with dimethyglyoxime. Find the percentages of Ni 
and Co in the ore. 

961. A nickel ore was analyzed by the volumetric method. The nickel 
solution was treated with KI solution and exactly 0.50 ml. of AgNO 3 solution 
containing 0.0125 gram of AgNO 3 per milliliter. The solution then reacted 
with 48.00 ml. of KCN solution containing 0.0140 gram of KCN per milliliter. 
What was the percentage of Ni in the ore if the sample taken weighed 0.900 
gram? 

962. What weight of dried glyoxime precipitate would be obtained from 
5.00 grams of steel containing 1.48 per cent Ni? 

Phosphorus 

(See also Probs. 256, 257.) 

963. How many milliliters of magnesia mixture (1.00 N with respect to 
MgCl 2 ) are required to precipitate the phosphorus from 0.2000 gram of pure 
apatite. Assume the formula of the latter to be 3Ca3(P0 4 )2.CaCl 2 ? How 
many grams of (NH 4 ) 3 PO4.12MoO 3 could theoretically be obtained from this 
weight of apatite? 

964. Calculate the percentage of phosphorus in a steel from the following 
data: 

Two grams of steel furnished a yellow precipitate which was dissolved in 



330 CALCULATIONS OF ANALYTICAL CHEMISTRY 

20.0 ml. of 0.500 N sodium hydroxide solution, and the excess of the latter 
reacted with 27.0 ml. of 0.333 N nitric acid. 

Calculate the weight of a sample of steel to be taken for analysis so that 
every 100 ml. of 0.100 N KMnO 4 used in the titration by the Blair method 
(see Part VI, under Phosphorus) will represent directly the percentage of P. 

Calculate in this process the equivalent weight of (a) P, (6) P 2 O 6 , (c) Mo, 
(d) MoO 3 , (e) Mo 24 O 3 7. 

966. A normal yellow precipitate of ammonium phosphomolybdate from 
a sample of bronze weighing 1.00 gram is reduced with zinc. The reduced 
solution requires 21.13 ml. of 0.100 N permanganate to oxidize the molyb- 
denum to the hexavalent condition. If the alloy contains exactly 0.20 per 
cent of phosphorus, to what hypothetical oxide was the molybdenum reduced 
by the zinc? 

966. What weight of steel should be taken for analysis so that the number 
of milliliters of 0.125 N permanganate required in the ferric alum method (see 
Part VI) will be two hundred times the percentage of P in the steel? 

967. A 2.00-gram sample of steel is dissolved in HNOa and the phosphorus 
is precipitated with molybdate as the normal yellow precipitate. The molyb- 
denum in the precipitate is reduced to a form corresponding to the oxide 
MoieO 2 7 and requires 10.0 ml. of 0.100 N KMnO 4 for reoxidation to the 
oxidation number of 6. Calculate the percentage of P in the steel. 

968. A sample is prepared for student analysis by mixing pure apatite 
[3Ca 3 (PO4)2.CaCl 2 ] with an inert material. If 1.000 gram of the sample gives 
0.4013 gram of Mg 2 P 2 O 7 , how many milliliters of ammonium oxalate solution 
[40.00 grams of (NH 4 ) 2 C 2 O 4 .H 2 O per liter] would be required to precipitate 
the calcium from the same weight of sample? 

969. Calculate the percentage of phosphorus in a sample of steel from the 
following data: 2.00 grams of steel furnished a normal yellow precipitate 
which when dissolved and passed through a Jones reductor reacted with 
7.00 ml. of 0.0833 N KMnO 4 . 

970. In the analysis of a sample of steel weighing 1.881 grams, the phos- 
phorus was precipitated with ammonium molybdate and the yellow precipitate 
was dissolved, reduced, and titrated with permanganate. If the sample 
contained 0.025 per cent P and 6.01 ml. of KMnO 4 were used, to what oxide 
was the molybdenum reduced? One milliliter of KMnO 4 was equivalent to 
0.007188 gram of sodium oxalate. 

971. The Pincus method for determining phosphate is to titrate it in acetic 
acid solution in the presence of ammonium ions with standard uranyl acetate 
solution according to the equation: PO^ + UO 2 + + + NH 4 + -> UO 2 NH 4 PO 4 . 
The indicator is ferrocyanide which gives a brown color on a spot plate with 
excess UO 2 ++. If the uranyl acetate solution is 0.100 N as an ordinary acetate 
salt and 10.0 ml. are used in the titration, how many grams of P 2 O 6 are shown 
to be present? 

972. A, solution containing phosphoric acid was treated with ammonium 
molybdate, and an abnormal yellow precipitate was obtained, which after 



PROBLEMS 331 



drying may be assumed to have consisted of [(NH^sPOJXMoOs)^ This 
precipitate was dried, weighed, and dissolved in ammonia water, and the 
solution was made up to 500 ml. Of this, 50.0 ml. were taken, made acid with 
H 2 S04, reduced with amalgamated Zn, and passed directly into an excess 
of ferric alum which served to oxidize the trivalent molybdenum back to 
the hexavalent condition. To oxidize the iron reduced by the molybdenum 
required a number of milliliters of 0.125 N KMnO 4 equal to 15.39 times the 
weight in grams of the original yellow precipitate. What values of x and y 
may be taken in the formula of the yellow precipitate? 

973. A carefully prepared steel is to be used as a standard for phosphorus 
determinations. It is analyzed by an accurate ("umpire") method in which, 
from a sample weighing 3.00 grams, a phosphomolybdate precipitate is 
obtained. This is dissolved and the phosphorus is subsequently precipitated 
as MgNH 4 PO 4 . On ignition this precipitate yields a pyrophosphate residue 
weighing 0.0109 gram. 

In the routine analysis of a plain carbon steel by the alkalimetric method, 
the analysis is run in parallel with a corresponding analysis of the above 
standard steel. The same weights of sample are used in the two cases, and a 
25-ml. pipetful of standard NaOH is used in each case. Back titration with 
HNO 3 using phenolphthalein indicator requires 10.2 ml. of the acid in the case 
of the standard steel and 8.6 ml. in the case of the unknown steel. If 1.00 ml. 
NaOH o 1.08 ml. HNO 3 , what is the percentage of phosphorus in the plain 
carbon steel? If the concentration of the HNOa were 0.105 N, what weight of 
sample must have been taken in each case? 

Ans. 0.0868 per cent. 3.0 grams. 

Sulfur. Selenium 

(See also Probs. 243, 244, 247, 251, 252, 265, 294, 373, 690, 696, 712, 714, 
729, 799, 1023.) 

974. A sample of ferrous ammonium sulfatc is prepared for student analysis 
by intimately mixing pure crystals of FeSO4.(NH 4 )2SO4.6H 2 O with an inert 
substance. Using a 0.7650-gram sample a student correctly obtains 0.1263 
gram of Fe 2 Os. What volume of barium chloride solution containing 25.00 
grams of BaCl 2 .2H 2 O per liter would be necessary to precipitate the sulfur 
from the filtrate? What is the percentage of inert material in the sample? 

976. A sample of pure ferric alum, FesCSOOs^NH^SO^HsO, is dissolved 
in water and the iron is precipitated with NH 4 OH. If the ignited precipitate 
weighs 0.1597 gram, (a) what volume of the NH 4 OH (sp. gr. 0.900, containing 
28.33 per cent NH 3 by weight) is theoretically required for the precipitation 
of Fe(OH) 3 , (6) how many milliliters of 0.1000 N BaCl 2 would be required to 
precipitate the sulfate from the iron filtrate, and (c) how many milliliters of 
0.1000 N Na 2 S 2 Os would be required in the determination of this amount of 
sulfate by the iodimetric (Hinman) method? 

976. If nitrogen is reduced to the 4-valent state, compute the volume 
of fuming nitric acid actually required to oxidize 5.000 grams of pyritic ore 



332 CALCULATIONS OF ANALYTICAL CHEMISTRY 

containing 70.10 per cent FeS2. Neglect the quantity of acid required for the 
remainder of the ore. Assume the acid to be of 1.500 specific gravity and to 
contain 94.10 per cent HNO 3 by weight. Also assume complete oxidation of 
sulfide to sulfate. Compute the weight of dry sodium peroxide required to 
carry out the same oxidation assuming the oxidation products to be NaFeO 2 
and Na 2 SO 4 . 

977. A sample of pure FeS 2 is analyzed by fusing a 0.5000-gram sample and 
precipitating the sulfur as BaSO 4 . How large an error in the weight of the 
precipitate must be made to produce an error amounting to 0.10 per cent of 
the apparent amount of S in the mineral? 

978. A soluble sulfate weighing 0.9261 gram is analyzed. The precipitate 
of BaSO 4 on ignition is found to weigh 1.3724 grams. On further ignition the 
weight increases to 1.3903 grams, owing to the fact that the precipitate as 
first weighed had been partly reduced to BaS which on further ignition was 
reoxidized to BaSO 4 . Calculate the true percentage of S in the original sample. 
Calculate the percentage of S present as sulfide and the percentage of S present 
as sulfate in the first ignition product. 

979. In the determination of sulfur by the evolution method, a notebook 
contains the following data: 

Weight of sample = 5.0275 grams 
Iodine used = 15.59 ml. 

Naa&Oi used = 12.68 ml. 

1.000 ml. iodine o 1 .086 ml. Na 2 S 2 O 3 
1.000 ml. Na 2 S 2 O 3 o 0.005044 gram Cu 

Find the percentage of sulfur. 

980. A steel weighing 5.00 grams is treated with HC1, and the evolved H 2 S 
is eventually titrated with a solution containing 0.0100 mole of KIO 3 and 0.4 
mole of KI per liter. What is the normality of the KIO 3 + KI solution as an 
oxidizing agent? If 3.00 ml. are used in the titration, what is the percentage 
of sulfur in the steel? 

981. A sample of steel contains 0.075 per cent sulfur. Using a 5.00-gram 
sample and determining the percentage of S in an analysis by the evolution 
method, calculate the molarity of KIO 3 to be used so that an error of 0.20 ml. 
in the titration will represent an actual error of only 0.001 per cent in the 
reported percentage of sulfur. 

982. A sample of A1 2 (S0 4 ) 3 .18H 2 which has lost a part of its water of 
crystallization and is therefore specified as A1 2 (SO 4 ) 3 .XH 2 O, is analyzed to 
determine its approximate composition. The calculation in this particular 
instance is based upon a determination of total sulfur as follows: A 0.5000-gram 
sample is dissolved in dilute hydrochloric acid and diluted, and the sulfate ion 
precipitated as BaSO 4 , yielding 0.5602 gram of ignited BaSO 4 . Calculate to 
three significant figures the value of X. 

983. It is desired to prepare a standard solution of iodine of such concen- 
tration that each milliliter will be equivalent to 0.010 per cent sulfur when the 



PROBLEMS 333 

latter is determined on a 5.00-gram sample by the evolution method. The 
iodine solution is to be prepared in the following way. A certain volume of 
0.105 N KMnO 4 is to be run from a buret into an aqueous solution containing 
an excess of KI, the solution is to be acidified with H 2 SO 4 and diluted to exactly 
1 liter. What volume of the KMnO 4 should be used? 
Ans. 297 ml. 

984. A carefully prepared steel is to be used as a standard in the evolution 
method for sulfur. A sample is dissolved in IINO 3 and the sulfur subsequently 
precipitated and weighed as BaSO4. The sample weighs 4.57 grams and the 
BftSO 4 weighs 0.0110 gram. 

In the routine analysis of a certain sample of Bessemer steel for sulfur by 
the evolution method the analysis is run in parallel with a corresponding 
analysis of the above standard steel. The same weights (5.00 grams) of sample 
are used. The above standard steel requires 3.3 ml. of KIO 3 + KI solution; 
the Bessemer steel requires 8.3 ml. What is the percentage of sulfur in the 
Bessemer steel and how many grains of KIO 3 does each liter of the titrating 
solution theoretically contain? 

Ans. 0.083 per cent, 1.12 grams. 

986. What volume (two significant figures) of 15.0 N HNO 3 would be used 
in dissolving the Cii2S precipitate from a 5.00-gram sample of steel containing 
0.25 per cent Cu, if the precipitate were contaminated with 5 per cent of its 
weight of FeS (assume sulfur completely oxidized to sulfate and the HNO 3 
reduced to NO 2 )? 

986. In the anlysis of 0.8000 gram of a substance for sulfur by the barium 
chromate method, 25.00 ml. of 0.1110 N sodium thiosulfate solution were used. 
Compute the percentage of sulfur. 

987. The H 2 S in a sample of illuminating gas is determined by passing 10.0 
cubic feet of the gas through an absorbing agent and oxidizing the sulfur to 
sulfate. By the Hinman method, there are used 12.00 ml. of thiosulfate 
solution having two-thirds the normality as a reducing agent as a certain 
potassium tetroxalate solution. 6.00 milliliters of the tetroxalate will reduce 
in acid solution 3.00 ml. of a KMnO 4 solution containing 0.00632 gram of 
KMnO 4 per milliliter. What is the H 2 S content of the gas in parts per thousand 
(by volume)? 

988. Sulfite liquor, used in the manufacture of sulfite paper pulp, consists 
essentially of a solution of Ca(HSO 3 ) 2 , Mg(HSO 3 ) 2 , and H 2 SO 3 . Titration with 
alkali converts all of these to normal sulfites (i.e., to SO 3 ~). Titration with 
iodine converts all sulfites and bisulfites to bisulfates. "Available S0 2 " is 
the actual free H 2 SO 3 plus one-half the SO 2 in the bisulfites of calcium and 
magnesium, and is given by the alkali titration (using phenolphthalein). 
" Combined SO 2 " is one-half the SO 2 in the bisulfites of calcium and magnesium 
and is given by subtracting the "available S(V from the "total SO 2 " (as 
given by the iodine titration). 

A 10-ml. pipetful of sulfite liquor (sp. gr. = 1.028) is introduced into a 
100-ml. measuring flask and diluted to the mark. A 10-ml. pipetful of the 



334 CALCULATIONS OF ANALYTICAL CHEMISTRY 

diluted solution is titrated with 0.03312 N NaOH, requiring 30.11 ml. to change 
the color of phenolphthalein. At this point starch is added, and the solution 
then requires 13.82 ml. of 0.1050 N iodine to give a blue coloration. 

(a) Calculate the percentage of "available SO 2 ," and of "combined SO*". 
(6) Calculate the percentage of "free SO 2 " (i.e., in the form of uncombined 
H 2 S0 8 ). (c) Show by an equation why the pink color produced at the first 
end point disappears at the beginning of the second titration. 

989. Fuming sulfuric acid is a clear, colorless, oily, fuming liquid, a mixture 
of the monohydrate (H 2 S04) and sulfuric anhydride (SOa) containing from 
13 to 15 per cent of free SOa. Fuming sulfuric acid for special purposes, com- 
mercially called oleum, contains as high as 60 per cent free SO 3 and is often 
partly or completely crystallized. The analysis depends upon the determina- 
ton of total acidity in a representative sample, carefully collected and weighed 
under conditions that assure no loss of material. The ordinary method of 
determining the acid strength does not take into account the effect of SOa 
which is always present in small quantities. When sulfurous acid, H 2 SO 3 , is 
titrated with base, methyl orange changes color when one hydrogen has been 
replaced; phenolphthalein changes color when both hydrogens have been re- 
placed. 

a. The analysis neglects the presence of S0 2 and assumes the mixture to 
be 1X2804 when 3.926 grams of oleum, dissolved in water, are diluted to exactly 
500 ml. A 100-ml. portion requires 34.01 ml. of 0.5132 N NaOH for complete 
neutralization, methyl orange being used as indicator. 

Calculate the percentage of H 2 SO4 and free SOa, expressing the answers 
to the number of significant figures justified by the data and method of 
calculation. 

b. The analysis includes the presence of SO 2 , and the mixture is calculated 
as H 2 SO 4 , SO 3 , and SO 2 . 

In this case, 3.926 grams of oleum dissolved in water are diluted to exactly 
500 ml. A 100-ml. portion requires 34.01 ml. of 0.5132 N NaOH for com- 
plete neutralization, methyl orange being used as indicator. Another 100-ml. 
portion is titrated directly with 0.1032 N iodine solution, starch being used as 
indicator, and requires 4.93 ml. 

Calculate the analysis in this case. 

990. An oleum contains only H 2 S0 4 , SO 3 , and S0 2 . The H 2 SO 4 and S0 3 
are found to be present in equal parts by weight. When a sample of the 
oleum is titrated with NaOH, phenolphthalein being used as indicator, the 
volume of alkali required is found to be fifty times the volume of iodine of 
the same normality required to titrate the SO 2 in the same weight of sample. 
Calculate the percentage composition of the oleum (see preceding problem). 

991. A sample of fuming sulfuric acid contains only H 2 SO 4 , SO 2 , and SOa. 
A sample weighing 3.2030 grams is dissolved in water and requires 5.00 ml. of 
0.2000 N iodine solution to oxidize the S0 2 . Another sample weighing 4.0301 
grams is titrated with 0.5000 N alkali with phenolphthalein as an indicator 
and requires 172.5 ml. What is the percentage composition of the acid, and 
what volume of alkali would have been used with methyl orange as the in- 



PROBLEMS 335 

dicator? (Methyl orange changes color when one hydrogen of H 2 SO3 has been 
replaced; phenolphthalein changes color when both hydrogens have been 
replaced.) 

992. To 5.00 ml. of a solution of a mixture of Na2S and NaHS (sp. gr. 1.032) 
is added a standard solution of HC1. The liberated H 2 S is determined by 
adding excess iodine solution and titrating back with thiosulfate, and the 
resulting acidity is determined with NaOH. From the following data, calculate 
the percentage of Na 2 S and the percentage of NaHS in the sample: 

1.000 ml. HC1 =0= 0.006005 gram CaC0 3 
12.0 ml. NaOH o 10.0 ml. HC1 

1.000 ml. I 2 o 0.00601 gram Sb 
10.0 ml. Na 2 S 2 O 3 o 12.0 ml. I 2 

HC1 used * 10.8 ml. 

NaOH used = 15.0 ml. 

I 2 used = 28.0 ml. 

Na 2 S 2 O 3 used = 15.0 ml. 

993. A solution that may contain Na 2 S, NaHS, H 2 S or mixtures of these is 
acidified and requires 20.00 ml. of 0.1000 N iodine solution for titration. The 
gain in acidity caused by the titration is equivalent to 10.00 ml. of NaOH 
(1.000 ml. =0= 0.006303 gram H 2 C 2 O 4 .2H 2 O). What does the solution contain? 
If 20.00 ml. of iodine had been required and the gain in acidity had been repre- 
sented by 17.00 ml. of NaOH, what would the solution have contained? 

994. A solution of a mixture of H 2 S and NaHS is acidified with a ml. of 
N/10 HC1, and the total H 2 S then present is determined by adding b ml. of 
N/10 I 2 and titrating back with c ml. of N/10 Na 2 S 2 O 3 . The acidity at 
the end of the titration is measured by d ml. of N/10 NaOH. Show that the 
number of grams of H 2 S present in the original solution is given by the formula 

[c + 2d - (6 + 2a)]0.001704 

995. The Norris and Fay method for determining selenium is to titrate 
with standard Na 2 S 2 O 3 according to the equation: H 2 SeO 3 + 4Na 2 S 2 O 3 + 
4HC1 -> Na 2 S 4 SeO 6 + Na 2 S 4 O 6 + 4NaCl + 3H 2 O. 

The Jamieson method for determining arsenic is to titrate with standard 
KIO 3 according to the equation: 2AsCl 3 + KIO 3 + 5H 2 O > 2H 3 AsO 4 + 
KC1 + IC1 + 4HC1. 

If the above KIO 3 is of such concentration that 3.00 ml. will liberate from 
excess KI in the presence of acid that amount of I 2 which reacts with 3.00 ml. 
of the above Na 2 S 2 O 3 , and 3.00 ml. of the Na 2 S 2 O 3 will react with 3.00 ml. of 
0.100 N I 2 , (a) what is the value of 1.00 ml. of the KIO 3 in terms of grams of As 
and (6) what is the value of 1.00 ml. of the Na 2 S 2 Os in terms of grams of Se? 

General and Miscellaneous Analyses 

996. A sample of pyrolusite analyzes as follows: MnO 2 = 75.00 per cent; 
CaO = 5.60 per cent; MgO = 4.00 per cent; Si0 2 = 15.40 per cent. A one- 
gram sample is dissolved in HC1 (Mn0 2 + 4HC1 - MnCl 2 + C1 2 + 2H 2 0), 



336 CALCULATIONS OF ANALYTICAL CHEMISTRY 

and the silica is removed in the regular way. The solution is neutralized and 
the manganese is precipitated with NH 4 OH + bromine water: MnCl 2 + 
Br 2 + 4NH 4 OH -> MnO 2 + 2NH 4 C1 + 2NH 4 Br + 2II 2 0. From the filtrate 
the calcium is precipitated as oxalate, and the precipitate is dissolved and 
titrated with KMn0 4 . The magnesium is precipitated with ammonium phos- 
phate in the regular way, and the precipitate is ignited and weighed. Calculate 
to 3 significant figures: (a) the number of milliliters of 3.00 N NH 4 OH and 
(6) the number of milliliters of 3.00 per cent Br 2 solution (sp. gr. 1.10) to pre- 
cipitate the manganese according to the above equation, (c) the total number 
of milliliters of 1.00 N Na 2 C 2 4 solution to form Mg(C 2 O 4 ) 2 = and precipitate 
all the calcium, (d) the number of milliliters of 0.100 molar KMnO 4 to titrate 
the precipitated calcium, (e) the weight of the ignited magnesium precipitate, 
(/) the percentage of Mn in the material obtained by strongly igniting a sample 
of the original pyrolusite in air, assuming conversion of MnO 2 to Mn 3 O 4 and 
no other changes. 

Ans. (a) 11.5 ml., (b) 41.8 ml., (<) 5.97 ml., (d) 4.00 ml., (e) 0.111 gram, 
(/) 52.2 per cent. 

997. A certain mineral has the following composition: CaO = 28.03 per 
cent; MgO = 10.05 per cent; FeO = 7.12 per cent; CO 2 = 44.77 per cent; 
giO 2 = 10.03 per cent. A one-gram sample is dissolved in HC1 and the silica 
removed in the regular way. Bromine is used to oxidize the iron. Calculate 
to 3 significant figures: (a) number of milliliters of 6.00 N HF theoretically 
required to volatilize the silica in the presence of concentrated H 2 SO 4 , (b) num- 
ber of milliliters of NH 4 OH (sp. gr. 0.960, containing 9.91 per cent NH 3 by 
weight) theoretically required just to precipitate all the iron as Fe(OH) 3 
after exact neutralization of the acid, (r) number of milliliters of KMnO 4 
(1 ml. =0= 0.00800 gram Fe 2 O 3 ) to titrate the oxalate in the regular calcium 
precipitate, (d) weight of ignited magnesium precipitate obtained in the usual 
way from the calcium filtrate. 

998. A certain mineral has the following composition: 



FeO = 14.41 per cent 
MnO = 7.12 per cent 
CaO = 28.00 per cent 
MgO = 3.89 per cent 
SiO 2 = 2.98 per cent 
C0 2 = 43.60 per cent 



A one-grain sample is decomposed without oxidizing the iron and put through 
a regular systematic analysis. Calculate to 3 significant figures: (a) number of 
milliliters of 6.00 N IIF theoretically required to volatilize the silica, (b) total 
number of milliliters of bromine water (sp. gr. 1.100, containing 3.00 per cent 
Br 2 by weight) and (c) total number of milliliters of 3.00 N NH 4 OH to pre- 
cipitate the iron and manganese together according to the equations: 2Fe++ + 
Br 2 + 6NH 4 OH~>2Fe(OH) 3 + 2Br- + 6NH 4 +; Mn++ + Br 2 + 4NH 4 OH -> 
MnO 2 + 2Br" + 4NH 4 ' f + 2H 2 0, (d) weight of this precipitate after ignition, 



PROBLEMS 337 

(e) total number of milliliters of 0.100 N H 2 C 2 O 4 solution to form the soluble 
complex Mg(C 2 O 4 ) 2 ra and completely precipitate the calcium, (/) weight of the 
material obtained by precipitating the magnesium with (NH 4 ) 2 HPO 4 in the 
regular way and igniting the precipitate, (g) number of milliliters of KMnO 4 
required to titrate the iron in a one-gram sample of the original mineral after 
decomposition without oxidation. Each milliliter of the KMnO 4 is equivalent 
to 0.006802 gram of NaCHO 2 in the following titration: 3CHO 2 ~ + 2MnO 4 - + 
H 2 O - 2MnO 2 + 3CO 2 + 5OII- 

Ans. (a) 0.333 ml., (b) 9.70 ml., (c) 3.33 ml., (d) 0.236 gram, (e) 140 ml., 
(/) 0.113 gram, (</) 6.00 ml. 

999. A sample of limestone rock was analyzed, and the percentages of the 
constituents, expressed as SiO 2 , Fe 2 O 3 , A1 2 O 3 , CaO, MgO, CO 2 , and H 2 O, were 
obtained as follows: 

SiO*. A sample weighing 2.500 grams was dissolved in IIC1, evaporated 
to dryness, and dried at 110C. The ignited insoluble residue weighed 0.6650 
gram, and all but 0.0015 gram was volatilized by 11F. This small residue was 
fused, dissolved, and added to the main filtrate. 

Fe'zOs + AkOz. The filtrate from the SiO 2 determination gave a precipitate 
with ammonia weighing 0.2181 gram after ignition. 

CaO. One-fifth of the filtrate from the above-mentioned determination 
gave a precipitate of calcium oxalate that required 38.40 ml. of 0.1225 N 
KMnO 4 for oxidation. 

MgO. From the filtrate of the Ca determination was obtained 0.1133 
gram of Mg 2 P 2 C>7. 

/^20s. A sample weighing 2.000 grams required 12.24 ml. of the above- 
mentioned KMnO 4 to oxidize all the iron after reduction. 

CO 2 + 7/ 2 0. A sample weighing 0.5134 gram after strong ignition weighed 
0.3557 gram. 

C0 2 . The same weight of sample on treatment with acid caused an ascarite 
bulb to gain 0.1541 gram in weight. 

What was the complete analysis as reported? 

1000. A mineral analyzes as follows: 

CaO = 23.9 per cent 

A/f~f\ Q 1 nt\t. f*f*-n4- 



M. 

MgO = 3.1 per cent 
Fe 2 O 3 = 40.0 per cent 
CO 2 = 33.0 per cent 



A one-gram sample is dissolved in HC1, evaporated to dryness, and taken 
up in just 10.0 ml. HC1 (sp. gr. 1.10, containing 20.0 per cent HC1 by weight). 
Calculate: (a) Total milliliters NH 4 OII (sp. gr. 0.97, containing 7.0 per cent 
NH 3 by weight) to neutralize the acid and just precipitate all of the iron. 
(6) Volume of KMnO 4 (having the same normality as a solution of thiosulfate 
of which 20.00 ml. will titrate the iodine liberated from excess KI by 0.1113 
gram of KBrO 3 ) to titrate the iron in the resulting precipitate after the iron 
has been dissolved in H 2 SO 4 and reduced with zinc, (c) Volume of KMn0 4 to 
titrate the calcium precipitate in the filtrate from the iron (1.00 ml. of this 



338 CALCULATIONS OF ANALYTICAL CHEMISTRY 

KMnO 4 o 1.00 ml. KHC 2 O 4 .H 2 C 2 O 4 .2II 2 O ~ 1.00 ml. NaOH o 1.00 ml. 
HC1 =c= 0.0106 gram Na 2 CO 3 ). (d) Weight of ignited Mg precipitate obtain- 
able from the Ca filtrate, (e) Percentage of Fe in the material obtained by 
completely igniting a large sample of the above material. (/) Empirical 
formula of the original mineral. 

1001. Ankerite is essentially a calcium-magnesium-ferrous carbonate occur- 
ring in nature as a vein mineral. A chemical analysis of a specimen of this 
mineral gave the following data: 

a. Moisture. A well-mixed sample was dried at 105C. to constant weight. 

Weight of sample = 10.000 grams 
Weight after drying = 9.988 grams 

b. Silica. A sample was dissolved in HC1, evaporated to dryness, heated 
at 105C. for one hour, and dissolved in IIC1, and the residue filtered, ignited, 
weighed, treated with HF, and reweighed. The small residue was fused with 
Na2CO 3 acidified and was added to the main solution. 

Weight of sample = 5.000 grams 
Weight of residue = 0.0417 gram 
Weight of residue after HF = 0.0117 gram 

c. d. Iron and Alumina. One-fifth of the filtrate from the silica determina- 
tion was treated with bromine and made ammoniacal, and the precipitate 
of ferric and aluminum hydroxides ignited and weighed. The ignited oxides 
were then fused with Na 2 CO 3 and dissolved in acid, and the iron reduced with 
Zn and titrated with KMn0 4 . Weight of oxides = 0.2845 gram; volume of 
0.1990 N KMnO 4 used = 17.33 ml. 

e. The filtrate from the combined oxides was treated with ammonium 
oxalate and the precipitated calcium filtered, dissolved in H 2 S04, and titrated 
with the above-mentioned KMn0 4 . Volume required = 47.50 ml. 

/. The filtrate from the calcium was treated with phosphate and the 
magnesium precipitate ignited. Weight ignited precipitate = 0.1430 gram. 

g. A 0.1000-gram sample of the original mineral was treated with HC1 
and the evolved CC>2 purified and caught in a special apparatus and measured 
over water previously saturated with C0 2 . 

Volume of CO 2 = 23.78 ml. 
Temperature = 20C. 
Barometric pressure = 747 mm. 
Vapor pressure of H 2 O at 20C. = 17 mm. 

Calculate percentage of: (1) H 2 O, (2) SiO 2 , (3) FeO, (4) A1 2 O 3 , (5) CaO, 
(6) MgO, (7) C0 2 . 

h. Neglecting the small percentages of.H 2 0, Si0 2 , and A1 2 3 , calculate an 
empirical formula for the mineral. 

i. How many milliliters of NH 4 OH (sp. gr. 0.96, containing 10,0 per cent 
NH 3 by weight) would be required in step c above to precipitate the iron 
and alumina after neutralization of the acid and boiling out of the excess 
bromine? 



PROBLEMS 339 

j. If a sample of the mineral were ignited in the absence of air so that all 
the II2O and C0 2 were lost, what would be the percentage of Fe in the resulting 
material? 

1002. A sample of manganese ore was submitted for analysis with the 
following instructions : 

Determine total silica, total manganese, total phosphorus, total iron, and 
oxidizing power. Previous experience with similar samples of this ore has 
indicated that the manganese may be present in two forms, Mn() 2 and Mn 3 O 4 , 
the phosphorus as phosphate, the iron as ferric iron, that no manganese is 
combined as silicate, and that the oxidizing power is due to MnO 2 as such or 
in the form of Mn 3 O 4 (MnO2.2MnO). Calculate: (a) total silica expressed as 
percentage of SiOo, (&) total iron as percentage of F(oO 3 , (r) total phosphorus 
as percentage of PsOs, (d) percentage of MnO 2 existing as such in the ore, 
(e) percentage of Mn 3 O 4 existing as such in the ore. 

Silica. Four samples weighing 2.000 grains each were fused with Xa^COa. 
Two fusions were dissolved in HNO 3 + H 2 O and two in TTC1. The solutions 
were evaporated arid the residues dehydrated. The first two were taken up 
in HNO 3 and the last two in HC1. Silica was determined in each case in the 
usual way, yielding 0.6260, 0.6274, 0.6268, 0.6268 gram, respectively. 

Total Manganese. The filtrate from the nitric acid silica determination 
was diluted to 1,000 ml. and 50-ml. portions used for manganese by the 
bismuthate method. In each case a 50-ml. pipetful of FeS0 4 (=0= 45.60 ml. 
0.1086 KMnO 4 ) was used, and the volumes of 0.1086 N KMnO 4 required in 
three determinations were 9.77, 9.75, and 9.79. 

Total Phosphorus. Two 100-ml. portions of the same solution were used 
for the alkalirnetric method. In each case a 25-ml. pipetful (o 29.30 ml. of 
0.09249 N HNO 3 ) of NaOH was added, and the titrations required 25.00 and 
25.04 ml. of 0.09249 N IINO 3 . 

Total Iron. The iron in the filtrate from the hydrochloric acid silica 
determination was reduced and subsequently required 2.00 ml. of 0.1086 N 
KMnO 4 for reoxidation. 

Total Oxidizing Power. This was determined by reduction of duplicate 
half-gram samples of the ore with 0.7000 gram of sodium oxalate in the 
presence of 1I 2 SO 4 . The excess oxalate required 27.00 and 27.06 ml. of 0.1086 
KMnO 4 . 

1003. A sample of moist lime sludge, a by-product from the manufacture 
of acetylene gas [CaC 2 + 2H 2 O - Ca(OH) 2 + C 2 H 2 ] was submitted for 
analysis, and the following results were obtained: 

Acid-insoluble Residue. A 2.000-gram sample treated with HC1 left a 
residue weighing 0.0274 gram. 

Combined Oxides (Fe 2 O 3 + A1 2 O 3 ). Treatment of the above-mentioned 
HC1 solution with NH 4 OH + NH 4 C1 gave a precipitate that ignited to 
0.0051 gram. 

Total Calcium. This was precipitated as oxalate from the above-men- 
tioned filtrate. The oxalate in the precipitate required 58.60 ml. of 0.4960 
N KMnO 4 . 



340 CALCULATIONS OF ANALYTICAL CHEMISTRY 

Total Magnesium. This was precipitated from the Ca filtrate as 
MgNH 4 PO 4 , yielding only a trace. 

Total Iron. A 4.000-gram sample of the original material was dissolved 
in HC1, reduced with Zn, and titrated with 0.1067 N KMnO 4 , the titration 
requiring 0.91 ml. 

Carbon Dioxide. A 10.00-gram sample of the original material, treated 
with HC1, evolved an amount of CO 2 to cause an ascarite absorption tube 
to gain 0.1031 gram in weight. 

Total Ignited Solids. A 3.000-gram sample of the original material, when 
dried and ignited, yielded a residue weighing 1.2704 grams. 

Total Alkalinity of Ignited Solids. The above-mentioned residue was 
dissolved in 100.0 ml. of 0.5927 N HC1 and diluted to 500 ml. and a 250-ml. 
portion titrated with 13.73 ml. of 0.5724 N NaOH. (This determination and 
that of total solids were made as a check and should be so used in the calcula- 
tion to prove the accuracy of the analysis.) (a) Calculate analysis of the moist 
lime .sludge in terms of percentages of: residue, Fe 2 3 , Al 2 Oa, calcium as 
Ca(OH) 2 and as CaCO 3 , and 1I 2 O (by difference), (b) Calculate analysis of 
the ignited residue. 

1004. The analysis of a sample of coal ashes produced by the combustion 
of a soft coal in a power station of a public-utility company was carried on by 
the following procedure with results of duplicate determinations as indicated. 
The analysis is similar to that of any undecomposable silicate. 

Loss on Ignition. The loss was taken at 800 to 900 C., representing ap- 
proximately the unburned fuel. Each sample weighed 20.000 grams. 

Loss: 3.0951, 3.0960 grams. 

Sulequent determinations were made on portions of the ignited material. 

Silica. One-gram samples were fused with sodium carbonate, the entire 
fusion dissolved in hydrochloric acid, and the silica determined by the usual 
method of dehydration. No correction was made with hydrofluoric; acid, as 
the sample Wifs not considered to be representative enough to warrant such a 
procedure. 

Weight of ignited SiO 2 : 0.5284, 0.5302 gram. 

Combined Oxides. The filtrate from the silica determination was used for 
the precipitation of combined oxides, assumed to be entirely ferric oxide and 
alumina. 

Weight of ignited oxides: 0.3927, 0.3920 gram. 

Calcium Oxide. The filtrate from the precipitation of the combined 
hydroxides was used for the precipitation of calcium as calcium oxalate, fol- 
lowed by the volumetric determination of the equivalent oxalate with 0.1020 
N KMn0 4 . 

Volume of KMnO 4 : 18.10, 18.05 ml. 

Magnesium Oxide. The filtrate from the precipitation of calcium oxalate 
was used to determine magnesia as MgNH 4 P0 4 .6H 2 O, ignited to Mg 2 P 2 O 7 . 

Weight of Mg 2 P 2 O 7 : 0.0257, 0.0256 gram. 

Total Iron. Two-gram samples were fused with sodium carbonate and 
dissolved in HC1, the silica was removed and the total iron determined 



PROBLEMS 341 

volumetrically with 0.1020 N KMnO4, after reduction with stannous 
chloride. 

Volume of KMnO 4 : 13.21, 13.21 ml. 

Total Sulfur. Two-gram samples were fused with sodium carbonate 
and dissolved in HC1, and silica and combined oxides were removed and the 
sulfate content was determined by precipitation as BaSO4. 

Weight of BaSO 4 : 0.0020, 0.0025 gram. 

Calculate: loss on ignition, SiO2, Fe 2 O 8 , A1 2 O 3 , CaO, MgO, SOa an the 
original sample. Report alkalies by difference. 

1006. The qualitative analysis of an alloy indicates the presence of small 
tin, large copper, small lead, small iron, small phosphorus, and large zinc. 
The quantitative data follow: 

Tin. The sample weighing 5.000 grams is treated with HNOs, evaporated 
to dryness, treated with HNOs, and filtered. The residue, ignited to SnO2 
and containing all of the phosphorus as P2C>6, weighs 0.0517 gram. 

Copper and Lead. The filtrate from the above is diluted to 500 ml., and 
a 50-ml. portion is electrolyzed. Cathode gains 0.2940 gram; anode gains 
0.0034 gram. 

Iron. The residual solution from the electrolysis is treated with NH 4 OII, 
the precipitated Fe(OH) 3 is redissolved, and the iron is reduced and titrated 
with 0.46 ml. of 0.1007 N KMnO 4 . 

Zinc. By difference. 

Phosphorus. A new sample weighing 2.000 grams is dissolved and the 
phosphorus eventually precipitated as MgNH 4 PO 4 . The ignited precipitate 
weighs 0.0131 gram. v 

Calculate the percentage composition of the alloy. 

1006. Given a sample of an alloy for the determination of tin, antimony, 
copper, and lead. Calculate the percentages of these four elements from the 
following data: 

Determination of Sb (Volumetric KMnO method). Sample = 1.000 gram. 
Volume of 0.1078 N KMnO 4 required = 17.70 ml. 

Determination of Sb + Sn (Residue from HNOs treatment). Sample == 
1.000 gram. Weight of ignited residue = 1.1478 grams. 

Determination of Cu and Pb (Electrolysis of filtrate). Sample = 1.000 gram. 
Increase in weight of cathode = 0.0428 gram. Increase in weight of anode = 
0.0583 gram. 

1007. The analysis of a sample of bronze of indefinite composition, but 
supposedly of a common commercial type, was carried out quantitatively 
without a preliminary qualitative analysis, as follows: 

Determination of Tin. A one-gram sample of the alloy gave a white 
residue in IINO 3 , indicating the presence of tin. This filtered residue was 
ignited and was found to weigh 0.0615 gram. 

Determination of Copper and Lead. The filtrate from the determination 
of tin was electrolyzed by the usual procedure. Gain in weight of cathode = 
0.8514; gain in weight of anode = 0.0512 gram. 

The fact that during the electrolysis no purple color developed at the 



342 CALCULATIONS OF ANALYTICAL CHEMISTRY 

anode, indicated the absence of manganese. The solution after electrolysis 
was slightly bluish, but continued electrolysis with clean electrodes gave no 
further deposition of metal. 

Determination of Iron. The solution and washings from the electrolysis 
were combined and the iron oxidized with bromine. Ammonia precipitated 
the characteristic ferric hydroxide, which gave an ignited product weighing 
0.001 1 gram. The filtrate was deep blue, showing the presence of nickel. 

Determination of Nickel. The regular dimethylglyoxime precipitate from 
the iron filtrate weighed 0.0205 gram. 

Determination of Zinc. The addition of ammonium sulfide to the nickel 
filtrate gave a white precipitate, showing the presence of zinc. This was 
redissolved and the zinc precipitated and weighed on a Gooch crucible as 
ZnNH 4 P0 4 . Weight = 0.1385 gram. 

Calculate and report the complete analysis. The sum of the percentages 
should serve as a check on the accuracy and completeness of the analysis. 

1008. A steel gave the following analytical data. Calculate the analysis. 
Carbon (Direct combustion method). Samples = 1.000 gram. CO2 ab- 
sorbed = 0.0100 gram. 

Manganese (Persulfate-arsenite method). Samples = 0.1000 gram. Titra- 
tion = 5.00 ml. The sodium arsenite solution used in the titration was pre- 
pared by dissolving exactly 0.3876 gram of pure As 2 O 3 in dilute Na2C0 3 and 
diluting to exactly one liter (assume reduction of the manganese to an average 
oxidation number of 3.50 in the titration). 

Phosphorus (Alkalimetric method). Samples = 2.000 grams. NaOH 
added = 25.00 ml. HNO 3 used = 26.91 ml. In the standardization of the 
solutions, 30.51 ml. of NaOH neutralized 0.8000 gram of KHC 8 H 4 O 4 , and 
25.00 ml. of the NaOH were equivalent to 33.19 ml. of the HNO 3 . 

Sulfur (Evolution method). Samples = 5.00 grams. Volume of KIO 3 
solution required = 3.05 ml. The KIO 3 solution was prepared by dissolving 
1.200 grams of KIO 3 and 12 grams of KI in water and diluting to exactly 
one liter. 

Silicon (Sulfuric acid dehydration). Samples = 1.000 gram. Weight of 
silica obtained = 0.0009 gram. 

Chromium (Persulfate method). Samples = 2.000 grams. Ferrous sulfate 
solution added = 25.00 ml. Titration required 17.75 ml. of KMnO 4 . In the 
standardization, 40.65 ml. of theKMnO 4 oxidized 0.2500 gram of Na 2 C 2 O 4 , and 
25.00 ml. of the ferrous solution were equivalent to 18.31 ml. of the KMnO 4 . 

1009. A sample of chrome-vanadium steel was analyzed, and the following 
mean values were obtained: 

Carbon (Direct combustion method). Sample = 1.000 gram. CO 2 ab- 
sorbed = 0.0176 gram. 

Manganese (Persulfate-arsenite method). Sample = 1.000 gram. Titra- 
tion required 3.85 ml. of arsenite. A standard steel weighing 0.100 gram and 
containing 0.66 per cent of manganese required 6.85 ml. of the same arsenite. 

Phosphorus (Alkalimetric method). Sample = 2.000 grams. NaOH 
added = 1 pipetful. Back titration required 31.4 ml. of HN0 3 of which 



PROBLEMS 343 

33.70 ml. were equivalent to one pipetful of the above NaOH. Normality of 
the HNO 3 = 0.09493. 

Sulfur (Evolution method). Sample = 5.00 grams. Titration required 
1.25 ml. of a standard solution of KIO 3 + KI of which 5.70 ml. were equivalent 
to the sulfur in 5.00 grams of Bureau of Standards steel containing 0.065 per 
cent sulfur. 

Silicon. Sample = 1.000 gram. Weight of Si0 2 after dehydration with 
II 2 SO 4 and ignition = 0.0334 gram. 

Chromium (Persulfate method). Sample = 2.000 grams. FeSO 4 added = 
1 pipetful. Back titration required 3.30 ml. of 0.1070 N KMnO 4 of which 
17.80 ml. were equivalent to one pipetful of the ferrous sulfate. 

Vanadium (Cr-V persulfate method). Sample = 2.000 grams. Volume of 
above KMnO 4 required = 1.30 ml. 

Calculate the percentage composition of the steel, and write equations 
for all fundamental reactions involved in the analysis, assuming the constit- 
uents to be Fe, Fe 3 C, MnS, Mn, Mn 2 Si 3 O 8 , F 3 P, Fe 2 Si, Cr, and V. 

1010. Two small duplicate samples of galvanized sheet were submitted 
for analysis with the statement that the product was manufactured by 
galvanizing copper-bearing steel by the usual process. The specifications 
called for the determination of the amount of zinc galvanizing calculated 
as ounces of metallic zinc per square foot of total surface, not including zinc 
on the edge of the sheet. They also called for the complete analysis of the 
steel that had been galvanized. 

The small pieces submitted were duplicate samples from the same sheet. 
An examination of the edges indicated that the samples had been cut from 
a larger sheet and therefore contained no zinc except on the two flat surfaces. 
In the analysis, the coating of zinc was removed by dipping the sheets in 
NaOH, and the zinc determined in the solution. The residual sheets of steel 
were combined as one sample and milled. Average values obtained in the 
analysis are given below. Calculate the results of the complete analysis 
according to specifications. 

ORIGINAL SHEET 

Sample 1 measuring 2% by 2% 6 inches and weighing 45.4409 grams. 
Sample 2 measuring 2% by 2% inches and weighing 46.9708 grams. 

ANALYSIS OF ZINC 

The caustic solutions were made acid, interfering constituents removed, 
and ^ aliquot portions taken. The zinc was determined in these aliquot 
portions by precipitation as ZriNH 4 PO 4 and weighing as such. Average 
weight of ZnNH 4 P0 4 in Sample 1 = 0.6393 gram; in Sample 2 = 0.6531 gram. 

ANALYSIS or STEEL 

Manganese (Bismuthate method). Sample = 1.000 gram. A 20.00-ml. 
pipetful of FeSO 4 was added, and the titration required 4.81 ml. of 0.1062 
N KMn0 4 (20.0 ml. FeSO 4 ^ 7.10 ml. KMnO 4 ). 



344 CALCULATIONS OF ANALYTICAL CHEMISTRY 

Phosphorus (Alkalimetric method). Sample = 2.000 grams. Volume of 
0.4617 N NaOH added = 5.00 ml. 

Titration with 0.1592 N HNO 3 required 11.92 ml. 

Silicon. A sample weighing 3.0 grams yielded 0.0008 gram of SiOa after 
dehydration with H 2 S(X 

Sulfur (Evolution method). Sample = 2.000 grams. A pipetful of iodine 
was eventually added to a solution of the evolved II 2 S and the excess was 
titrated with 16.31 ml. of 0.1001 N Na 2 S 2 O 3 . A pipetful of the iodine was 
found to be equivalent to 18.08 ml. of the thiosulfate. 

Carbon. Combustion of a one-gram sample of the steel formed 0.0039 
gram of CO 2 . 

Copper. Precipitated as CuS, redissolved, and determined iodimetrically. 
Volume of above thiosulfate for 10.0-gram sample = 1.04 ml. 

1011. A sample of meat scrap is submitted for analysis. The material 
consists principally of a mixture of beef and bone that has been processed by 
heating, and the sample has been ground in a Wiley mill to a fairly fine con- 
sistency. Material of this type is used commercially as an important com- 
ponent of poultry food, dog biscuit, and similar products. 

The scrap from which the sample was taken was sold under the following 
specifications: 

Protein: not less than 45 per cent 

Ash: not greater than 35 per cent 

Bone phosphate: within the limits of 25 to 33 per cent 

Grease: not greater than 10 per cent 

Free fatty acid: not greater than 10 per cent of the grease 

Moisture: not greater than 9 per cent 

Crude fiber: not greater than 2 per cent 

The following numerical data represent the averages of duplicate determina- 
tions in each case. Calculate the analysis of the material as indicated. Does 
it conform to specifications? 

Protein. A 2.000-gram sample was analyzed by the Kjeldahl method 
(see Part VI, under Nitrogen). The evolved NII 3 was caught in a 5 per cent 
solution of boric acid and titrated with standard IIC1, requiring 19.40 ml. 
The IIC1 was standardized against the NII 3 liberated from pure (NH 4 ) 2 SO 4 
[1.000 ml. =0= 0.03490 gram (NH 4 ) 2 SO 4 ]. Arbitrary factor for converting 
percentage of nitrogen to percentage of protein = 6.25. 

Moisture. A sample weighing 5.000 grams was dried to constant weight 
at 105 C. Weight of dried material = 4.638 grains. 

Ash. The material from the moisture determination was ignited at dull 
red heat. Weight of residue = 1.611 grams. 

Bone phosphate. This means phosphate expressed as Ca 3 (PO 4 ) 2 . The ash 
obtained above was dissolved in HNO 3 , the solution evaporated dry, and the 
residue taken up in dilute HNO 3 . The solution was filtered and a Ko aliquot 
portion was treated with (NH 4 ) 2 Mo0 4 . The yellow phosphomolybdate 
precipitate was filtered and dissolved in NH 4 OH, and the phosphate was then 



PROBLEMS 345 

precipitated as MgNH 4 PO 4 and ignited. Weight of Mg 2 P 2 O 7 = 0.0250 
gram. 

Grease. A 3.000-gram sample of the original material was dried and ex- 
tracted with anhydrous ether for 8 hours. The ether extract was evaporated. 
Weight of residue = 0.2700 gram. 

Crude Fiber. The grease-free material was digested with dilute H 2 SO 4 and 
then with dilute NaOH according to exact specifications of procedure. The 
residue was filtered off on an alundum crucible and dried at 105C. Weight 
of residue (== fiber + inorganic material) plus crucible = 11.8366 grams. 
The crucible plus residue was then ignited at dull red heat. Weight of residue 
(inorganic material) plus crucible == 11.8016 grams. 

Free Fatty Acid. The grease from the above ether extraction was heated 
with alcohol and titrated with standard NaOH, using phenolphthalein indi- 
cator. Volume of 0.05050 N NaOH required = 2.16 ml. Free fatty acid is 
usually expressed as percentage of oleic acid (milliequivalent weight = 0.282) 
present in the grease rather than in the original material. 

Ans. Protein = 44.85 per cent. Moisture = 7.24 per cent. Ash = 32.22 
per cent. Bone phosphate = 27.87 per cent. Grease = 9.00 per cent. Crude 
fiber =1.17 per cent. Free fatty acid = 11.40 per cent. 

1012. A sample of solution submitted for analysis is known to contain 
chromium sulfate, potassium dichromate, and free sulfuric acid. All other 
constituents are eliminated by information as to the source of the solution. 
Preliminary experiments indicate that the content of the various constituents 
to be determined is such that different volumes of the original must be used 
as samples. Fifty milliliters of the original are diluted to exactly 500 ml. 
and portions used for the analysis as specified. 

Determinatimi of Total Chromium Content. The chromium in 50.00 mi. 
of the diluted solution is oxidized by ammonium persulfate in acid solution, 
a measured excess of standard ferrous sulfate is added, and the excess titrated 
with standard permanganate solution. 

25.00 ml. FeSO 4 o 23.38 ml. KMn0 4 
75.00 ml. FeSO 4 solution added 
7.93 ml. 0.1020 N KMnO 4 used in titrating the excess 

Determination of Trivalent Chromium. The chromium is precipitated 
as Cr(OH) 3 from a 100.00-ml portion of the diluted solution and the precipitate 
ignited to Cr 2 O 8 . Ignited precipitate = 0.1623 gram. 

Determination of Total Sulfate. The filtrate from the above-mentioned 
trivalent chromium determination is diluted to exactly 250 ml., and a 
100.00-ml. portion is used to determine total sulfate gravimetrically as 
BaSO 4 . Weight of ignited BaSO 4 = 1.132 grams. 

Calculate: (a) Grams of Cr 2 (SO 4 ) 3 per 100 ml. of original sample, (b) grams 
of K 2 Cr 2 O 7 per 100 ml. of original sample, (c) amount of free H 2 SO 4 expressed 
in terms of acid normality. 

1013, A sample of impure potassium perchlorate was submitted for an 
analysis for chloride and chlorate content. On the assumption that only 



346 CALCULATIONS OF ANALYTICAL CHEMISTRY 

these three acidic constituents were present, and only as potassium salts 
(which was justified by a review of the method of preparation of the original 
material), the following analysis was made: 

The chloride present was precipitated from dilute HNO 3 solution with 
AgNO 3 (chlorate and perchlorate are not precipitated). Weight of sample = 
5.000 grams. Weight of silver chloride = 0.0118 gram. 

In another sample, the chlorate present was reduced in neutral solution 
by prolonged heating with ferrous sulfate (perchlorate is not reduced). After 
dissolving of the precipitated basic ferric salts in HN0 3 , the total chloride in 
the solution was precipitated with AgNOa. Weight of sample = 5.000 grams. 
Weight of silver chloride = 0.0501 gram. 

Calculate the percentages of KC1, KC1O 3 , and KC1O 4 (by difference). 
Calculate the amount of FeSO4.7H 2 O necessary to reduce the chlorate present. 

1014. It is proposed to discharge the spent dye liquor from a dyehouse, 
amounting at times to 126 gallons per minute, into a neighbouring stream. 
Laboratory tests indicate that this may be done satisfactorily if the volume of 
the stream is sufficient to dilute the dye liquor one thousand times. Tests of the 
stream flow are made by adding to the stream a solution of sodium chloride 
at the rate of one gallon in 24 seconds. The chloride in the stream above the 
point of dosing is found by titrating 100* ml. with 0.01156 N AgNO 3 , 1.10 ml. of 
the silver solution being required. A 100-ml. sample taken below the point of 
dosing required 1 .22 ml. of the same solution. Each milliliter of the dosing solu- 
tion required 73. 17 ml. of AgNO 3 . (a) What is the stream flow in gallons per min- 
ute? (b) What is the normal chloride content of the stream in parts per million? 
(c) What dilution would be obtained for the maximum discharge of dye liquor? 

Ans. (a) 153,000 gallons per minute. (6) 4.51 parts per million, 
(c) 1,214 times. 

1015. Most samples of leather when moistened are acid to litmus. This 
reaction, however, unless extremely marked and in the presence of much 
sulfate, is not conclusive evidence of free mineral acid. Although there is 
no simple chemical method that will give an accurate estimate of this acid, it 
is possible to compare different leathers for acidity by a method of Proctor 
and Searle (Leather Industries Laboratory Book of Analytical and Experimental 
Methods, page 371, 1908). 

The principle of this method follows: A sample of finely divided leather 
is treated with standard sodium carbonate solution and evaporated, the 
leather carbonized, and the residue leached with water. The carbonization 
drives off the organic sulfur without an appreciable reduction of sulfur to 
sulfide. The solution is filtered, and the residue ashed and treated with 
standard hydrochloric acid. This solution is mixed with the original filtrate and 
the mixture titrated with standard alkali, methyl orange being used as indicator. 
The data on an actual determination follow: 

Sample weight 5.000 grams 

Na 2 C0 3 added 25.00 ml. 

HC1 added 28.50 ml. 

Titration, NaOH 14.50 ml. 



PROBLEMS 347 

The sodium hydroxide is standardized against 0.8000 gram of potassium 
acid phthalate, requiring 34.25 ml. If 25.00 ml. of the Na 2 CO 3 0= 28.50 ml. 
HC1, calculate the free mineral acid as percentage of H 2 SO 4 by weight in the 
sample. 

1016. A sample of tanned sole leather was submitted for analysis with 
specification that the following determinations be made: 

1. Ash on original. 

2. Total chromium on ash, calculated to percentage of Cr 2 Oa in the ash 
and percentage of Cr 2 0a in the original. 

3. Total sulfur in the original, calculated as percentage of S. 
The analysis and data obtained were as follows: 

1. Duplicate 4.000-gram samples were incinerated at a dull red heat, 
until the carbonaceous matter was consumed. 

Weight of ash = 0.3912, 0.3915 gram 

2. Duplicate 0.1500-gram samples of the ash were fused with a mixture 
of Na 2 C0 3 and Na 2 O 2 , the fusion leached with water, the solution boiled to 
remove excess peroxide and made acid, excess standard ferrous sulfate added, 
and the excess ferrous sulfate titrated with standard KMnO 4 . 

Volume of FeSO 4 added = 100.00 ml. 
Normality of KMnO 4 = 0.1057 
Ratio: 50.00 ml. FeSO 4 ~ 24.60 KMnO 4 
Back titratioii, KMnO 4 = 4.00; 3.98 ml. 

3. Duplicate 1.000-gram samples were mixed with pure Na 2 C0 3 , the mixture 
covered with Na 2 C0 3 , and incineration carried on at a temperature below the 
fusion point of Na 2 CO 3 to prevent loss of sulfur compounds. The residue was 
leached with dilute HC1 and the sulfate precipitated as BaSO 4 , after the re- 
moval of silica. 

Weight of ignited BaS0 4 = 0.0755; 0.0747 gram 

Calculate as indicated. 

1017. Commercial zinc dust, or "blue powder/' is a fine, gray powder 
obtained as a by-product from the production of zinc and is generally com- 
posed of zinc (80 to 90 per cent), zinc oxide (9 to 10 per cent), lead (1.5 to 
2.0 per cent), and traces of cadmium, iron, arsenic, and antimony. The 
amount of zinc oxide depends partly upon the care with which oxygen is 
excluded during storage. As this material is often used industrially as a 
reducing agent it is usual to determine its reducing power with solutions of 
K 2 Cr 2 O 7 . 

A 1 .600-gram sample of the powder, 100.00 ml. of 0.5000 N K 2 Cr 2 O 7 , and 
10 ml. of 6 N sulfuric acid are shaken in a bottle with additions of small 
amounts of acid until the sample is dissolved. The solution is then diluted 
in a measuring flask to exactly 500-ml., a 100-ml. portion is taken, 5 ml. HC1 
and 10 ml. of 10 per cent KI solution are added, and the liberated I 2 is titrated 
with 12.96 ml. of 0.1252 N thiosulfate solution, starch being used as indicator. 



348 CALCULATIONS OF ANALYTICAL CHEMISTRY 

Calculate the reducing power of this sample as percentage of zinc. Which of 
the metals listed above contribute to the reducing power of the sample? 

1018. Bleaching powder (chloride of lime) when treated with water forms 
calcium hypochlorite [Ca(OCl) 2 ] and calcium chloride. The calcium hypo- 
chlorite constitutes the active bleaching and disinfecting agent, and the analysis 
is therefore a determination of " available chlorine/' expressed as percentage 
of Cl by weight, which is the chlorine present as hypochlorite, but not the 
chlorine present as chloride. The original material must be kept in airtight 
containers and protected from air as far as possible, since bleaching powder is 
acted upon by carbonic acid, which liberates hypochlorous acid with a corre- 
sponding loss of available chlorine. Representative samples for analytical 
work are difficult to obtain and handle; but the material itself is inexpensive, 
and results obtained are sufficiently accurate for the purposes for which the 
material is used. 

A 10.00-gram sample is triturated with successive small portions of water 
until it is well ground. The portions and residue are washed into a liter 
measuring flask and thoroughly mixed, and an aliquot of 50.00 ml. taken for 
analysis. The sample should carry its proper proportion of sediment, for 
the supernatant liquid gives percentages below, and the sediment percentages 
above, the average. The sample is slowly titrated with sodium arsenite 
solution (4.425 grams pure As 2 O 3 , 13 grams Na 2 CO 3 diluted to exactly 1,000 ml.) 
until a drop of the solution gives no color to iodostarch paper as an outside 
indicator (OC1~ + AsO 3 ^ -> Cl + As0 4 "). 

From the data given in the problem, prove that the number of milliliters 
of the arsenite used, multiplied by 2, gives the number of liters of chlorine 
(at OC. and 760 mm.) per kilogram of substance. This corresponds to the 
chlorometric (or Gay-Lussac) degree. 

Prove also that the chlorometric degree must be multiplied by 0.31698 to 
give the American degree (percentage of Cl by weight). 

1019. A method for the determination of chloride, bromide, and iodide 
in the presence of each other, devised by Julius Bekk [Che?n. Ztg., 39: 405-406 
(1915); Chem. Abstracts, 9:2042 (1915)] as a result of research on this separa- 
tion, invites the following procedure: 

1. The halogens are precipitated and determined in total as silver salts. 

2. A similar mixture of washed, moist silver halides precipitated from a 
second sample is digested with 2 grams of potassium dichromate in 30 ml. 
of concentrated sulfuric acid at 95 C. for 30 minutes. The iodide is oxidized 
to iodic acid; the chloride and bromide are liberated as chlorine and bromine 
and removed by passing a stream of air through the solution. The solution 
is diluted and filtered to remove anhydrous insoluble chromic sulfate and 
the iodic acid reduced to iodide by adding drop by drop a concentrated solution 
of sodium sulfite until a liberal excess is present. (Sulfurous acid causes the 
separation of iodine until an excess has been added.) The precipitated silver 
iodide is determined by filtration and ignition. 

3. The filtrate from (2) containing the silver formerly with the chloride and 
bromide is determined as silver iodide by precipitation with a soluble iodide. 



PROBLEMS 349 

Write all equations involved in these reactions. 

In a given case the qualitative analysis of the original substance shows that 
Na and K are the only positive constituents. From the following data calculate 
the milliequivalents of Cl, Br, and I in the original mixture. Calculate the 
milliequivalents of K in the sample, and by inspection of your answer calculate 
the most probable percentage composition of the original. 

(a) Weight of sample = 0.5000 gram 

Precipitated halides = 1.0643 grams 

(6) Weight of sample = 0.5000 gram 

Precipitated Agl = 0.07074 gram 

(c) Precipitated Agl = 1.5044 grams 

(d) Weight of sample = 0.5000 gram 
KC1O 4 = 0.04176 gram 

1020. Lactic acid as a laboratory preparation can be produced by boiling 
cane sugar or glucose with a solution of sodium hydroxide. The acid can be 
separated by converting it into the calcium salt which can then be purified 
by crystallization and reconverted to the acid by treatment with H 2 SO4. 
The acid is monobasic and has the formula CH 3 .CHOH.COOH. 

A sample of the calcium salt produced in this way is analyzed by ignition 
of the salt to CaO, which may be weighed directly, titrated with standard 
acid, or checked by both methods. A one-gram sample of the calcium salt 
gives an ignited residue weighing 0.2481 gram. This residue is then titrated 
by adding 50.00 ml. of 0.5132 N IIC1 and titrating back with 32.38 ml. of 
0.5194 N NaOH. 

Calculate the purity of the calcium lactate by each of the two methods. 

1021. Sodium bismuthate is a yellow, or brown, amorphous, somewhat 
hygroscopic powder, not a true bismuthate but of variable composition, 
containing about 4 per cent of "active" or available oxygen, corresponding 
to 70 per cent NaBiOa. It is insoluble in cold water but is decomposed by 
boiling water. The salt is used principally in the determination of man- 
ganese. 

Three grams of sample were mixed with 20 grams of KI and 30 ml. of 
water in a 100-ml. graduated flask. Twenty milliliters of dilute HC1 were 
added, the flask was well shaken, stoppered, and allowed to stand for one hour 
in the dark. The solute was diluted to the mark and mixed and an aliquot of 
20.00 ml. taken for analysis. The titration required 29.42 ml. of 0.1022 N 
thiosulfate, starch being used as indicator. 

Write the equation for the fundamental reaction. Calculate the percentage 
of available oxygen in the material. 

1022. The following data are taken from the chemical analysis of a solution 
containing copper sulfate and free sulfuric acid only. The color of the solu- 
tion and a preliminary test for acidity indicate the presence of a large amount 
of copper and a slight degree of acidity. The color prevents the use of a 
conventional volumetric acid measurement. 

Determination of Copper. A 25.00-ml. sample of the original solution is 



350 CALCULATIONS OF ANALYTICAL CHEMISTRY 

diluted to 250 ml. in a graduated flask and thoroughly mixed, and 50.00 ml. 
taken for the analysis by the iodimetric method. A volume of 33.72 ml. of 
thiosulfate (33.35 ml. =0= 0.2241 gram Cu) is required. 

Determination of Total Sulfate. A 25.00-ml. portion of the original solution 
is diluted to 250 ml. in a graduated flask and thoroughly mixed, and 50.00 ml. 
taken for analysis. BaSO 4 obtained = 1.170 grams. Calculate: (a) normality 
of the thiosulfate solution; (6) grams of copper per liter of solution; (c) grams 
of free sulfuric acid per liter; (d) normality of the solution as an acid; (e) milli- 
liters of 6.00 N sulfuric acid to be added to a liter of the solution to make it 
exactly 0.6000 N as an acid. 

1023. "Bisulfite liquor" is an aqueous solution of calcium and magnesium 
bisulfites [Ca(HSO 3 )2 and Mg(HSO 3 ) 2 ] and excess free sulfurous acid. It is 
made by passing SO 2 gas through a suspension of Ca(OH) 2 and Mg(OH) 2 and 
generally contains a small amount of sulfate because of the presence of S0 3 in 
the gas. The liquor is used in the sulfite digestion process for the production 
of paper pulp; it disintegrates the wood chips by rendering the noricellulose 
parts soluble. 

For control tests in the mill, a volumetric method is usually sufficient. 
Gravimetric methods are used for a complete and more precise analysis. 

Determination of Specific Gravity. By means of a Westphal balance deter- 
mine the specific gravity of the liquor. Value obtained = 1.050. 

VOLUMETRIC CONTROL ANALYSIS 

Into a 100-ml. volumetric flask transfer a 10-ml. pipetful of the liquor, 
dilute to the mark, and mix. 

Determination of Total 80%. Titrate with standard iodine a 10-ml. pipetful 
of the above prepared solution. Volume of 0.1010 N I 2 required = 10.05 ml. 
(IISO 3 - + I 2 + H 2 O -^ HSOr + 21- + 2H+). 

Determination of Available S0 2 . Titrate with standard NaOH (using 
phenolphthalein) a 10-ml. pipetful of the prepared solution. Volume of 0.1 100 
N NaOH required = 6.04 ml. (IlSOr + OH- -> SOr + H 2 O). 

GRAVIMETRIC ANALYSIS 

Determination of Silica. Evaporate a 25-ml. pipetful of the original liquor 
with HC1 to dryness. Dehydrate, dissolve in HC1, filter, and ignite residue in 
the regular way. Weight of residue = 0.0027 gram. 

Determination of Fe^Oz + AW*. Use the filtrate from the silica determina- 
tion and precipitate with NII 4 OII. Filter and ignite in the regular way. 
Weight of ignited precipitate = 0.0051 gram. 

Determination of CaO and of MgO. Evaporate a 25-ml. pipetful of the 
original liquor with H 2 SO 4 to dryness. Weight of CaSO 4 + MgSO 4 = 0.5875 
gram. Dissolve in HC1, add NH 4 OH and (NH 4 ) 2 C 2 4 . Filter the precipitated 
CaC 2 O 4 and ignite. Weight of CaO = 0.2225 gram. 

Determination of SO*. Pipet 100 ml. of the original liquor into a flask, add 
HC1, and boil out the SO 2 in a current of CO 2 to exclude air. Precipitate the 
sulfate with BaCl 2 . Weight of BaS0 4 = 0.0330 gram. 



PROBLEMS 351 

DISCUSSION ANT> CALCULATIONS 

"Available SO 2 " is the free H 2 S0 3 plus one-half the SO 2 in the calcium and 
magnesium bisulfites and indicates the SO 2 in excess of the amount necessary 
to form neutral sulfites. It is given by the titration with NaOII. " Total S0 2 " 
is given by the iodine titration. " Combined SO 2 " is represented by one-half 
the SO 2 in the bisulfites of calcium and magnesium and is found by subtracting 
"available SO 2 " from "total SO 2 ." 

(a) From the volumetric analysis calculate the percentages of "available 
SO 2 ," "total SO 2 " and "combined SO 2 ." (6) From these values find the per- 
centage of "free SO 2 " (i.e., as free H 2 SO 3 ). (c) From the gravimetric analysis 
calculate the percentages of SiO 2 , Fe 2 O 3 + A1 2 O 3 , and S0 3 . (d) From the 
gravimetric analysis of calcium and magnesium calculate the weight of SO 2 
combined as Ca(HSO 3 ) 2 and Mg(HSO 3 ) 2 . One-half of this is " combine<l SO 2 ." 
Calculate this percentage and compare with the value obtained volumetrically. 
Ans. (a) 2.03 per cent, 3.10 per cent, 1.07 per cent; (6) 0.96 per cent; 
(c) 0. 10 per cent, 0.019 per cent, 0.011 per cent; (d) 0.5581 gram, 1.063 per cent. 

1024. Commercial aluminum sulfate is used in the sizing of paper, as a 
mordant, in water purification, and for various other purposes. It is some- 
times contaminated with iron and may be " acidic " (containing free H 2 SO 4 ) 
or "basic" (containing free A1 2 O 3 ). The following is an analysis of a typical 
sample: 

Insoluble Matter. A 25.00-gram sample was dissolved in hot water and 
filtered. The residue, dried at 105C., weighed 0.0525 gram. The filtrate 
was diluted to exactly 500 ml. and mixed (Solution 1). Of this solution, 
100 ml. were taken and diluted to exactly feOO ml. in another measuring flask 
(Solution 2). 

Total Iron and Aluminum. A 50-ml. pipetful of Solution 2 was made acid, 
the iron oxidized with HNO 3 , and the iron and aluminum precipitated with 
NII 4 OH + NI1 4 C1. Weight of ignited precipitate = 0.07602 gram. 

Total SO S . A 100-ml. portion of Solution 2 was made acid with HC1 and 
BaCl 2 added. Weight of ignited precipitate = 1.0250 grams. 

Total Iron. A 100-ml. portion of Solution 1 was made acid with II 2 SO 4 , 
poured through a Jones redurtor, and titrated with 0.05272 N KMriO 4 , of 
which 1 .37 ml. were required. 

Acidity or Basicity. The method [/. Soc. Chem. Ind., 30: 184 (1911)] 
is based upon the following reaction: 

A1 2 (SO 4 ) 3 + (H 2 SO 4 ) + 12KF -> 2(A1F 3 .3KF) + 3K 2 SO 4 + (H 2 SO 4 ) 

Potassium fluoride solution, titrated to neutrality (phenolphthalein being 
used) with acid or base, decomposes aluminum salts forming stable com- 
pounds reacting neutral to phenolphthalein; but any free acid that may be 
present remains as such. Of Solution 1, 68.00 ml. were taken, diluted, and 
heated to boiling, and 10.99 ml. of 0.5176 N H 2 SO 4 were added. An excess 
of KF solution was added to the cooled solution, and 10.00 ml. of 0.5292 N 
KOH were required to neutralize the solution to phenolphthalein. 

It is customary to calculate the iron to FeS0 4 even though some of it may 



352 CALCULATIONS OF ANALYTICAL CHEMISTRY 

exist in the ferric state. The remaining SO 3 is calculated to A1 2 (SO 4 ) 3 . If 
the sample is "basic," the A1 2 O3 left over is reported as A1 2 O 3 . If the sample 
is acidic, the SO 3 left over is reported as free H 2 SO 4 . 

Express the results of the analysis in terms of percentage of insoluble 
matter, FeSO 4 , A1 2 (SO 4 ) 3 , "basic aluminum" (A1 2 O 3 ) or "free acid" (H 2 SO 4 ), 
and water (by difference). 

1026. Lime mortar, made from sand and lime, hardens by taking up carbon 
dioxide, the lime being changed to calcium carbonate. Unlike hydraulic 
mortar or cement, it will not harden unless fairly dry and exposed to the 
air. 

Cement mortar is a mixture of Portland cement, sand, and water. Cement 
mortar is much harder and more durable than lime mortar, but some lime is 
often added to make it spread and work more readily. The proportion of 
sand to cement is not greater than 3:1. Cement mortar is more expensive 
than common mortar but is useful in masonry that is exposed to water or 
where great strength is required. The setting of cement mortar is a process 
of hydration. The analysis of mortar to obtain the composition of the original 
material must be made on a sample after ignition. 

A sample of mortar that has been in use "and therefore has absorbed CO 2 
or has been set by chemical action, or both, is analyzed in duplicate in 
accordance with the following procedure. 

The original material which is known to be representative is crushed gently 
and thoroughly mixed. A portion of indefinite weight large enough to use as 
a sample for all determinations is ignited. 

Sand. A weighed portion is treated with a considerable volume of dilute 
HC1 arid warmed, the solution decanted, and this process repeated until all 
the soluble matter has been removed. Though some silicic acid precipitated 
by the acid may be included in the residue, the amount is small and is com- 
pensated to some extent by the iron and alumina dissolved from the sand. 
The residue is filtered off and ignited as "sand." The filtrate is diluted to a 
definite volume and an aliquot taken that is small enough to allow the precipita- 
tion of a reasonable amount of calcium and magnesium. 

Elimination of Silica. By evaporation to dryiiess in accordance with the 
usual method. Precipitate discarded. 

Elimination of Combined Oxides (Fe 2 3 , A1 2 O 3 ). By precipitation with 
ammonia. Precipitate discarded. 

Determination of Calcium. By precipitation as CaC 2 O 4 and titration with 
KMnO 4 in the presence of H 2 SO 4 . 

Determination of Magnesium. By precipitation as MgNH 4 P0 4 and igni- 
tion to Mg 2 P 2 O 7 . 

Calculate and interpret the analysis from the data obtained as shown below. 
These are mean values from the duplicate determinations. 

The interpretation of this analysis is based upon the usual assumption 
that the average sample of Portland cement contains 62 per cent CaO and 
that the ignited material consists entirely of sand, cement, and free lime 
(with MgO). 



PROBLEMS 353 

Sample for sand = 5.000 grams 

Residue (sand) = 3.9869 grams 

Dilution of filtrate = 500.0 ml. 

Portion taken = 100.0 ml. 

Normality KMnO 4 = 0.1317 

Volume of KMnO 4 used = 37.67 ml. 

Weight of Mg 2 P 2 O 7 = 0.05521 gram 

1026. Structural concrete consists of cement, sand, and rock to which 
water has been added in proper amount to bring out a process of hydration. 
The chemical analysis of concrete calls for a determination of the proportions 
by volume of the original mix previous to the addition of the water and may 
be exceedingly complicated or comparatively simple depending upon the 
extent to which data are required. The simplest type of analysis, in which 
the sand and rock are actually determined and the cement is assumed to be 
the difference, is illustrated in the following problem. 

A representative sample of material in wliich the cement adheres loosely 
to the rock and in which it is apparent that the concrete has failed because 
of improper original mixture, or other factors, is weighed and the entire sample 
disintegrated with a hammer or mortar and pestle, care being taken to knock 
the cement off the coarse aggregate or stone without breaking or crushing the 
stone or sand particles. The rock particles (those which will not pass a j^-inch 
sieve) are weighed as "rock." The finer portion is thoroughly mixed and 
sampled, and the loss on ignition is determined on a small portion. Another 
small portion is treated with dilute hydrochloric acid to dissolve the cement, 
and the sand is collected by filtration and is ignited and weighed as such. 

DETERMINATION ON ORIGINAL CONCRETE SAMPLE 

Weight of original concrete = 1,785 grams 
Weight of rock = 575 grains 

DETERMINATIONS ON ROCK-FREE SAMPLE 
Sand: 

Weight of sample = 15.00 grams 

Weight of sand = 12.15 grams 
Loss on ignition: 

Weight of sample = 5.000 grams 

Ignition loss = 0.2654 gram 

a. Calculate on a percentage-by-weight basis the loss on ignition and the 
sand on the rock-free sample. 

b. Determine the cement by difference. 

c. Convert this 100 per cent analysis on the rock-free sample to a 100 per 
cent basis on the original concrete by introducing the percentage of rock as 
a factor. 

d. Convert this 100 per cent analysis of the original concrete to a 100 per 
cent analysis on a water-free and CCVfree basis by eliminating loss on ignition 
as a factor. 



354 CALCULATIONS OF ANALYTICAL CHEMISTRY 

e. On the generally accepted basis that 

1 cubic foot of rock = 100 pounds 
1 cubic foot of sand = 90 pounds 
1 cubic foot of cement = 94 pounds 

convert the analysis from a percentage-by-weight to a parts-by-volume basis. 
/. Reduce these data to a unit basis for cement to two significant figures. 

1027. The complete qualitative analysis of a sample of welding compound 
showed definitely the presence of Na+, K+, Ol~, and F~, the first three con- 
stituents present in large amounts in comparison with the fluoride. The 
analysis also proved conclusively the absence of all other constituents, 
including water in any form. 

The following quantitative analyses were made: 

The determination of total Cl~, precipitated and weighed as AgCl 

Sample weight = 0.4000 gram 
AgCl weight = 0.7794 gram 

The determination of total F~, precipitated and weighed as CaF2. 

Sample weight = 3.000 grams 
CaF 2 weight = 0.3046 gram 

The determination of total Na+ and K+ (after removal of the interfering 
constituent F~) by the usual perchlorate method. 

Sample weight = 0.2000 gram 

NaCl and KC1 weight = 0.2095 gram 
KC1O 4 weight = 0.1714 gram 

A calculation of the percentage of the constituents K+, Na+, Cl*~, and F~ 
gives no indication of the proportions of the salts actually mixed to make this 
compound. An interpretation of the results to give this information should 
be made in accordance with the following method: 

a. Calculate the weights of all precipitates, and convert to the basis of a 
one-gram sample. 

b. Reduce all data to the milliequivalent basis. 

1. Milliequivalents calculated from weight AgCl. 

2. Milliequivalents calculated from weight CaF 2 . 

3. Milliequivalents calculated from weight KCKX 

(1) Convert milliequivalents KC1O 4 to grams KC1 

(2) Subtract grams KC1 from grams KC1 + NaCl 

(3) Calculate milliequivalents from weight NaCl. 

c. Total the milliequivalents due to + constituents. 

d. Total the milliequivalents due to constituents. 

If these totals balance within the limits of experimental error, it indicates 
that a correct analysis has been made, barring the improbable possibility 
that a constituent of the compound was completely overlooked in the qualita- 
tive and quantitative analyses. * 



PROBLEMS 355 

e. Consider the seven possible mixtures of all the salts of the constituents 
K+, Na+, Cl~, F~, starting with the simpler mixtures, as enumerated 
below*: 















KCl 


KCl 


KF 


KF 


KF 


KCl 


KCl 


NaCl 


NaF 


NaCl 


NaF 


NaF 


NaCl 


NaCl 


KF 






NaCl 


KCl 


NaF 


KF 


NaF 



/. By balancing the milliequivalents it is possible to eliminate four of the 
first six mixtures listed as impossible from the data of the analysis and to 
prove two possible mixtures as a duplication of the sample. 

The possibility of the seventh mixture cannot be proved or disproved with 
the data of the analysis. Further examination, possibly of a microscopic 
or complicated chemical nature, would be necessary to indicate the actual 
existence of the four separate salts. 

g. From the observations made in (/), calculate the percentage composition 
of the mixture, bearing in mind that all data are now based on a one-gram 
sample. The percentage mixture is an interpretation of the actual salts and 
their proportions by weight to be used in duplicating the original welding 
compound. 

Ans. KCl = 46.09 per cent KF = 15.11 per cent 

NaCl = 43.43 per cent KCl = 26.70 per cent 

NaF = 10.91 per cent NaCl = 58.61 per cent 

1028. The qualitative analysis of a solution of brine used in a special 
refrigerating process indicates the presence of barium, potassium, and sodium 
in medium amounts and chloride in large amount. All other constituents 
are eliminated. The method of quantitative analysis used in this particular 
case, which specified an inexpensive commercial analysis, is as follows: 

Specific Gravity. This was determined with a hydrometer at 20 C. = 1.188. 

Total Barium. A 50.00-ml. portion of the original solution was diluted 
to exactly one liter, and a 100.00-ml. portion was used for the precipitation 
of barium as BaSO 4 . Weight ignited BaSO 4 = 0.6169 gram. 

Total Chloride. A 50.00-ml. portion of the original solution was diluted 
to exactly one liter, and a 50.00-ml. portion was used for the precipitation 
of chloride as AgCl. Weight AgCl = 1.2575 grams. 

Total Solids. Twenty-five milliliters of the original solution were evapo- 
rated to dryness and ignited at a temperature of 400C to remove the water 
of crystallization in the BaCl2. The residue (assumed to be BaCl 2 , NaCl, 
KCl) weighed 6.7912 grams. 

Calculate the percentage by weight of BaCl 2 .2H 2 0, NaCl, and KCl in 
the original solution. 

1029. Acetaldehyde is a low-boiling liquid which is soluble in water. 
The dilute aqueous solution when treated with a solution of sodium bisulfite 
forms an addition compound as indicated by the following equation: 
CH 3 CHO + NaHSO 3 + H 2 O - CH 3 CHOHSO 8 Na. It was desired to work 



356 



CALCULATIONS OF ANALYTICAL CHEMISTRY 



out a feasible iodimetric method for determining acetaldehyde making use of 
this reaction. Since both bisulfite ion and the above addition product under 
proper conditions of acidity can be oxidized by iodine, and since bisulfite in 
the presence of acid loses S(>2, a satisfactory method would seem to require 
careful control of such factors as pit value and time of standing. The experi- 
mental analyses tabulated below were conducted in an attempt to use such an 
iodimetric method. In each analysis a 25.0-ml. pipetful of a solution contain- 
ing 3.324 grams of acetaldehyde (mol. wt. = 44.0) and a little H 2 S0 4 per liter 
was used. The aldehyde employed was carefully prepared and was better 
than 99 per cent pure. To this aldehyde solution, after dilution with water, 
was added the indicated quantity of a solution containing approximately 
39.5 grams of NaHS0 3 in 3.5 liters. After standing the indicated time, the 
solution was titrated with 0.0926 N iodine, with starch as the indicator. In 
certain cases (as indicated below) IIC1 was added together with the bisulfite 
and in certain cases (as indicated) NaHC0 3 was added after reaching the 
end point with iodine. When NaHC0 3 was used, it was found that further 
iodine was necessary to restore the blue color, the total volume required being 
indicated in the last horizontal columns. The titrations were conducted in 
Erlerimeyer flasks and no special precautions were taken to eliminate loss of 
SOa on standing. 

Convert all numerical data to inilliequivalents and from a study of the 
values obtained write out in detail a dependable method for determining 
acetaldehyde to a precision of two or three significant figures. Emphasize 
those conditions that must be followed with care and formulate the method 
of calculating the results. State all the conclusions you can draw as to the 
chemical characteristics of the reactions involved. 





1 


2 


3 


4 


6 


NallSOs soln 


35.0 ml. 


35.0 ml. 


35.0 ml. 


35.0 ml. 


35.0 ml. 


2NHC1 


None 


None 


None 


None 


None 


Time of standing 


5 min. 


15 min. 


30 min. 


10 min. 


15 min. 


First 1 2 end point 


68.6 ml. 


64.4 ml. 


59.2 ml. 


10.9 ml. 


18.2 ml. 


NaHC0 3 added 


None 


None 


None 


Excess 


Excess 


Second I 2 end point 








35.2 ml. 


50.5 ml. 

















G 


7 


8 


9 


10 


NaHSOa soln 


35.0 ml. 


35.0 ml. 


35.0 ml. 


35.0 ml. 


35.0 ml. 


2 N HCi 


1.0 ml. 


1.0 ml. 


1.0 ml. 


1.0 ml. 


1.0ml. 


Time of standing 


5 min. 


15 min. 


20 min. 


30 min. 


60 min. 


First 1 2 end point 


34.6 ml. 


33.6 ml. 


26.8 ml. 


24.2 ml. 


29.2 ml. 


NaHCOg added 


Excess 


Excess 


Excess 


Excess 


Excess 


Second I 2 end point . . . 


64.5 ml. 


69.9 ml. 


64.4 ml. 


64.8 ml. 


70.0 ml. 



PROBLEMS 



357 





11 


12 


13 


14 


15 


NaHSOs soln 


25.0 ml. 


50.0 ml. 


35.0 ml. 


35.0 ml. 


35.0 ml. 


2N HC1 


1.0 ml. 


1.0 ml. 


1.0 ml. 


1.0 ml. 


1.0 ml. 


Time of standing 


30 min. 


30 min. 


30 min. 


30 min. 


30 min. 


First 1% end point 


21.6 ml. 


53.1 ml. 


31.6 ml. 


36.7 ml. 


29.9 ml. 


NaHCO 3 added 


Excess 


Excess 


Until 


Excess 


Excess 


Second I 2 end point . . . 


58.8 ml. 


93.8 ml. 


neutral 
69.5 ml. 


77.3 ml. 


70.6 ml. 



1030. " Niter cake" (commercial sodium acid sulfate), a by-product from 
one of the processes for the manufacture of nitric acid, is used extensively 
as an acid in the pickling of steel and other alloys. The usual impurity is 
a small amount of iron as ferric sulfate. A sample is submitted for analysis 
and the following information requested: (a) the percentage of total iron, 
calculated to percentage Fe 2 (SC>4)3, (b) the acid strength of the sample cal- 
culated as percentage of NaHS0 4 , (c) any free acid calculated to percentage 
of H2SO 4 , (d) any deficiency of acid calculated as percentage of Na 2 SO 4 . 

Determination of Total Iron. The iron was precipitated as Fe(OH)s, 
dissolved, reduced, and titrated with 0.1050 N KMnO 4 . Sample = 10.00 
grams. Volume of KMnO 4 required = 0.20 ml. 

Determination of Total Sulfate. Precipitated as BaSO 4 (without elimina- 
tion of small iron) from a hydrochloric acid solution of the salt. Sample = 
2.000 grams. Solution diluted to 250 ml. and a 50-ml. aliquot portion taken. 
Ignited BaSO 4 obtained = 0.7760 gram. * 

Determination of Total Add Strength. Direct titration with 0.5137 N 
NaOH on a two-gram sample, methyl orange being used as the indicator, 
required 32.02 ml. of the NaOH. 

a. Calculate all data to a one-gram basis, and reduce to milliequivalents, 
being careful to designate the milliequivalents in the iron titration as obtained 
by a process involving oxidation and reduction, in the case of BaSO 4 as a 
sulfate, and in the alkali titration as an acid. 

6. Convert the number of milliequivalents due to the iron content, which 
lias been calculated as an oxidizing agent, over to the number of milliequiva- 
lents as a salt (sulfate), and convert the milliequivalents obtained from the 
alkali titration over to milliequivalents as a sulfate. The results should 
indicate an excess sulfate as Na2S0 4 . 

c. Calculate the milliequivalents of iron to percentage of Fe2(SO 4 )s. 
Subtract the milliequivalents of iron expressed as a salt from the total 
milliequivalents of BaSO 4 , and calculate the milliequivalents of NaHSO 4 as 
an acid or as a salt to percentage of NaHS0 4 . 

d. Calculate the residual milliequivalents of BaSO 4 to percentage of Na2S0 4 . 

1031. A deposit taken from a steam boiler, a typical "boiler scale," had 
the following characteristics: dark-colored scales, apparently containing 
iron oxide; the center layer of the plates white in color, indicating the possi- 
bility of CaS0 4 ; general characteristics indicating some organic matter. The 



358 CALCULATIONS OF ANALYTICAL CHEMISTRY 

specification of analysis called for a typical analysis of the boiler scale and 
indicated that the sample contained nothing uncommon to such a product. 
The specification also called for an interpretation of results. The following 
analyses were made: 

Sampling. The scale was broken up in a porcelain mortar, then quartered 
and a final portion ground to a fine powder in agate. 

ANALYSIS OF OKIGINAL SAMPLE 

Moisture. A portion of the original sample was dried at 105 C. to constant 
weight. 

Sample weight = 9.0000 grains 

Weight of dried sample = 7.6303 grams 

ANALYSIS OF THE DRIED SAMPLE 

Oil. A portion of the dried sample from the moisture determination was 
transferred to an extraction thimble and extracted with ether in a Soxhlet 
extractor in the usual way, the extract being collected in a weighed flask. 
The ether was distilled off and the flask dried at 105C. 

Weight of sample = 3.0000 grams 

Weight of flask and extract = 17.3296 grams 
Weight of flask = 17.3027 grams 

ANALYSIS OF THE ORIGINAL SAMPLE 

Organic and Volatile Matter. A portion of the original sample was ignited 
at a low temperature until the organic matter was burned off. The tempera- 
ture was sufficient to remove moisture, oil, organic matter, and C0. from 
carbonates but insufficient to decompose sulfates. 

Weight of sample = 1.0000 gram 
Weight of residue = '0.7286 gram 

ANALYSIS OF THE IGNITED SAMPLE 

Insoluble in Acid (Siliceous material). A portion of the ignited sample 
was treated with dilute HC1 until only a white silica residue remained; the 
residue was filtered, washed, ignited, and weighed in the usual way. 

Weight of sample = 1.0000 gram 

Weight of ignited residue = 0.5370 gram 

Iron and Aluminum Oxides. The filtrate from the silica determination 
was oxidized with HNO 3 , NH 4 C1 and NH 4 OH added, the NIT 4 OH nearly 
all expelled by boiling, and the precipitated A1(OH) 3 and Fe(OH) 3 filtered 
and ignited. Weight of ignited oxides = 0.2280 gram. 

Calcium. The filtrate from the combined oxides precipitation was used 
to precipitate CaC 2 O 4 , which was filtered, dissolved in H 2 SO 4 , and titrated 
with KMn0 4 . 

Volume of KMnO 4 = 27.40 ml. 
Normality of KMnO 4 = 0.1070 



PROBLEMS 359 

Magnesia. The filtrate from the calcium determination was used for 
precipitation of MgNH 4 PO 4 in the usual way. The precipitate was ignited. 
Weight of ignited precipitate = 0.03181 gram. 

Sulfur Trioxide. A 1.000-gram portion of the ignited material was treated 
with concentrated HC1, diluted, and filtered and the sulfate precipitated as 
BaSO 4 and ignited. Weight of BaSO 4 = 0.4089 gram. 

Chlorine. A portion of the ignited sample was treated with water and 
filtered and the chloride content determined in the filtrate by the usual 
method of precipitating AgCl in the presence of HNO 3 , filtering, and drying 
at 105C. Weight of AgCl = none. 

Total Iron. A portion of the ignited sample was dissolved in HC1, the 
iron reduced with stannous chloride, the excess reducing agent reoxidized 
with HgCl 2 , and the iron titrated with KMn(>4 in accordance with the usual 
method. 

Weight of sample = 1.0000 gram 
Volume of KMnO 4 = 26.77 ml. 
Normality KMnO 4 = 0.1070 

Carbon Dioxide. A portion of the original substance showed no effer- 
vescence with dilute HC1. 

1. Using the data obtained from the analysis of the ignited sample, which 
is all on a one-gram basis, calculate on the ignited sample: (a) percentage by 
weight of insoluble residue; (b) percentage by weight of combined oxides 
(Fe 2 O 3 .Al 2 O 3 ); (c) grarn-milliequivalents of CaO; (d) gram-milliequivalents 
of MgO; (e) gram-milliequivalents of S0 3 ; (/) gram-milliequivalents of Cl; 
(#) percentage by weight of Fe expressed as* Fe 2 O 3 ; (h) gram-milliequivalents 
of CO 2 . 

2. Calculate: (a) milliequivalents Cl (if present) to percentage of NaCl 
by weight; (6) milliequivalents SO 3 to percentage of CaSO 4 if an excess of 
SO 3 is present, to percentage of MgSO 4 if SO 3 is insufficient to combine with 
CaO, remainder of CaO to percentage of CaCO 3 ; (c) milliequivalents MgO 
in excess of the excess milliequivalents SO 3 , to percentage of MgO (not to 
percentage of MgCO 3 , for this decomposes at the temperature of the boiler). 

3. Consolidate the analysis on the ignited sample to percentages of 

Insoluble residue 
Iron oxide (Fe 2 O 3 ) 
Alumina (A1 2 O 3 ) 
Calcium sulfate (CaSO 4 ) 
Magnesium sulfate (MgSO 4 ) 
Magnesium oxide (MgO) r 

Sodium chloride (NaCl) 

(This analysis of the ignited portion, if correct, should total 100 per cent 
within the limits of experimental error.) 

4. Calculate the percentage of organic and volatile matter (including oil, 
moisture, organic matter, and C0 2 ) on the original sample, and convert the 
100 per cent analysis on the ignited sample to a 100 per cent analysis on an 
original basis by introducing this factor. 



360 CALCULATIONS OF ANALYTICAL CHEMISTRY 

5. Calculate the moisture content on the original sample, and introduce it as 
a factor in the analysis, deducting the percentage for the "organic and volatile 
matter" and changing this term to "organic matter + oil + CO 2 ." 

6. Calculate the percentage of C0 2 in the original sample, and introduce 
it as a factor in the analysis, deducting the percentage for the "organic 
matter + oil + CO 2 ," leaving now a factor "organic matter + oil." 

7. Calculate the percentage of "oil in the dried sample," and convert 
this percentage only to an original sample basis. Deduct this value for the 
factor "organic matter + oil," leaving the factor "organic matter." 

8. Make out a complete report of analysis on the original sample basis. 
Ans. Moisture = 15.22 per cent 

Oil = 0.76 

CO 2 = none 

Organic matter = 11.16 
Insoluble in acid = 39.14 
Fe 2 O 3 = 16.68 

Al 2 Os = none 

CaSO 4 = 14.53 

MgSO 4 = 2.50 

MgO = none 

NaCl = none 

99.99 per cent 

1032. Two samples of table or dairy salt, submitted for complete analysis, 
were analyzed in accordance with the method given below, with results as 
indicated. The analysis was made in duplicate by the standard methods 
to determine whether these samples conformed to the United States specifica- 
tions which require that they shall contain, on a water-free basis, not more 
than the following amounts of impurities: 

1.4 per cent CaS04 
0.5 per cent CaCl 2 + MgCl 2 
0.1 per cent insoluble in H 2 O 
0.05 per cent BaCl 2 

(In addition to the substances specified, table salt sometimes contains 
small amounts of Ca 3 (PO 4 ) 2 , Na 2 SO 4 , and MgS0 4 . Natural salt also may 
contain small amounts of Na 2 C0 8 , KC1, and other minor impurities.) 

METHOD OF ANALYSIS AND RESULTS 

Appearance. Examine the material under a microscope and note its general 
appearance. It should be homogeneous and free from foreign matter. Add a 
drop of dilute HC1 to the salt on the slide, and note if there is any effervescence 
due to the presence of carbonates. 

Sample 1: Homogeneous, clear color, no foreign matter. Definite crystal- 
line structure. No effervescence with HC1 

Sample 2: Homogeneous, clear color, no foreign matter. Definite crystalline 
structure. Grains much smaller than Sample 1. No effer- 
vescence with HC1 



PROBLEMS 361 

Solubility and Reaction. Make a nearly saturated solution with distilled 
water. Test with sensitive litmus paper. A turbidity which dissolves on the 
addition of HC1 indicates CasCPO^ or CaCO 8 . 
Sample 1 : Clear solution, neutral to litmus 
Sample 2: Turbid solution, neutral to litmus; turbidity disappears on the 

addition of acid. 

Moisture. Dry 10 grams to constant weight at 105C. 
Sample 1: 

Sample 10.0000 10.0000 

Wt. dry 9.8603 9.8592 

Loss 0.1397 0.1408 



Sample 2: 
Sample 
Wt. dry 
Loss 


10.0000 
9.5520 


10.0000 
9.5480 


0.4480 


0.4520 



Phosphoric Anhydride. Dissolve 50.00 grams in distilled water, dilute 
to 500 ml., and pipet out 100-ml. portions of uniform solution and suspension. 
Add 10 ml. of concentrated HNO S ; then add NH 4 OH until the acid is nearly, 
but not completely, neutralized. Add an excess of ammonium molybdate 
solution, warm gently, and let stand one hour. If the solution is colored 
bright yellow but gives no precipitate, report a trace of P 2 Os. If a yellow 
precipitate forms, determine the phosphate by one of the standard methods 
(see Part VI, under Phosphorus). 
Sample 1: 

Sample = 10.00 grams 
Analysis = no precipitate obtained 
Sample 2: 

Sample = 10.00 grams 
KMnO 4 = 30.25 ml. \ m . . , , 
= 30.10 ml. } Blairmeth d 
KMnO 4 = 0.1067 N. 

Iron Oocide and Alumina. (a) In the absence of P 2 O 6 : To a new 100-mL 
portion of the above-mentioned solution add a few drops of concentrated 
HNO 3 , and boil to oxidize the ferrous iron. Add a slight excess of NH 4 OH, 
filter, wash, and ignite to FeuOa + A1 2 O 8 . (fc) In the presence of P 2 O 5 : To 
100 ml. of the original solution, add slightly more than enough ferric sulfate 
to combine with the P 2 C>5. After adding the iron solution, add a slight excess 
of NH 4 OH, and boil until barely ammoniacal. This precipitates all the P 2 Os 
as FePO 4 and excess iron as Fe(OH) 8 . Filter, wash, and ignite to Fe 2 O 8 + 
A1 2 O 8 + P 2 5 . From this weight, subtract the amount of P 2 O 6 previously 
determined and the weight of Fe 2 O 8 added. The remainder will be Fe 2 8 + 
A1 2 8 . 

Sample 1: 

Sample weight = 10.00 grams 
Ignited precipitate = none 



362 CALCULATIONS OF ANALYTICAL CHEMISTRY 

Sample 2: 

Sample weight = 10.00 grams 

Milliliters Fe 2 (SC>4)3 solution added = 100.0 
(0.001275 molar) 

T . . , . . . . \ 0.0269 gram 

Igmted precipitate = | 00269 * 

Total Calcium. In the filtrate from the combined oxides, precipitate the 
calcium as CaC204.H 2 O, dissolve in dilute H 2 SO4, and titrate the equivalent 
oxalate with standard KMnC>4. 

Sample 1: 

Milliliters KMnO 4 = { ^'j[J 

KMnO 4 = 0.1067 N 

Sample 2: 

Milliliters KMn0 4 = j j'^ 

KMnO 4 = 0.1067 N 

Total Magnesium. Precipitate the magnesium in the filtrate from the 
calcium oxalate as MgNH 4 PO 4 .6H 2 O; ignite to Mg 2 P 2 O 7 . 

a i i T -A j Ti/r ^ ^ f 0.01751 gram 
Sample 1: Ignited Mg 2 P 2 O 7 = | Q Q1757 B 

a i o T M j ** r rt \ 0.07296 gram 

Sample 2: Ignited Mg 2 P 2 7 = | Q 073(X) * 

Sulfur Trioxide. To 100 ml. of the original solution, add 5 ml. of dilute 
HC1 and precipitate BaSC>4 in the usual way. 

o i i TIT T.J. -n cn-v f 0.2446 gram 
Sample 1 : Weight BaSO 4 = | Q 244g * 



Sample 2: Weight BaSO 4 = { JJjjg gram 



Total Barium. If sulfate is present, Ba cannot be in the solution but 
might be present in the insoluble portion. If SO 3 was not found, test for 
barium by adding 5 ml. of dilute H 2 SO 4 to 100 ml. of the original solution, 
precipitating BaSO 4 , igniting in the usual way. 

Total Chloride. Dilute 100 ml. of the original solution to 500 ml., mix, 
and pipet out 25 ml. Dilute to 500 ml., add 5 ml. of dilute HNO 3 , and pre- 
cipitate as AgCl, drying at 105C. 

Sample 1: Weight AgCl = { |;JjgJ * rams 
Sample 2: Weight AgCl = { };}JJJ grams 

Total Potassium. Dilute 100 ml. of the original solution to 150 ml. Heat 
to boiling, and add, drop by drop with constant stirring, a slight excess of 
BaCl 2 solution. Without filtering, add in the same manner Ba(OH) 2 solution 
in slight excess. Filter while hot, and wash until free from chloride. Add 
to the filtrate 1 ml. of concentrated NH 4 OH and a saturated solution of 
until the excess barium is precipitated. Heat, and add 0.5 gram 



PROBLEMS 



363 



of oxalic acid; filter, wash until free from chloride, evaporate the filtrate to 
dryness in a platinum dish, and ignite carefully over a free flame below red 
heat until all volatile matter is driven off. Digest the residue with hot water 
and filter through a small filter. Acidify with HC1, and add H 2 PtCl 6 solution 
in excess. Evaporate on a water bath; add 80 per cent alcohol. Filter on a 
Gooch crucible, and dry at 105C. 

Sample 1: 
K 2 PtCl 6 = none 

Sample 2: 

K 2 PtCl 6 = 0.4304 gram 

a. Calculate loss at 105 C. directly to percentage of loss by weight. 

6. Calculate grams of P 2 O 6 , grams of Fe 2 O 3 + A1 2 O 3 + P 2 O 6 , grams of 
Fe 2 O 3 + A1 2 3 , grams Fe 2 3 , and grams A1 2 3 per gram basis; reduce the 
single terms to milliequivalents. 

c. Reduce all other data to milliequivalents per gram basis. 

d. Calculate: P 2 O 6 to Ca 3 (PO 4 ) 2 ; excess over CaO to Mg 3 (P0 4 ) 2 ; further 
excess to Na 2 HPO4. 

e. Calculate: SO 8 to CaSO 4 ; excess over CaO to MgS0 4 ; further excess to 
Na 2 SO 4 . 

/. Calculate CaO over P 2 O 6 and SO 3 to CaCO 3 (if the salt solution is turbid 
and shows the presence of carbonates) or to CaO (if the salt is alkaline) or to 
CaCl 2 (if the solution is clear and neutral). 

g. Calculate MgO over P 2 O 6 and SO 3 to MgCO 3 (if the salt solution is 
turbid and shows presence of carbonates) or to MgO (if the salt solution is 
alkaline) or to MgCl 2 (if the solution is clear and neutral). 

h. Report Fe 2 O 3 and A1 2 O 3 as such. 

i. If BaCl 2 , CaCl 2 , MgCl 2 , KC1 are present, subtract the equivalent 
amount of AgCl from the total AgCl before calculating the latter to NaCl. 

j. Calculate percentage by weight of each constituent of the original 
sample. 

k. Calculate percentage by weight of each constituent (dry basis). 

Ans. 





Sample 1 


Sample 2 


Ca 3 (PO 4 ) 2 


None 


0.150 


Fe 2 O 3 


None 


None 


A1 2 O 3 


None 


None 


CaCl 


0.324 


0.054 


MgCl 2 


0.15 


0.653 


CaSO 4 


1.45 


2.65 


BaO 


None 


None 


NaCl 


98.17 


95.00 


KC1 


None 


1.39 








Total 


100.09 


99.90 









APPENDIX 



TABLE IV. DENSITY OP WATER AT TEMPERATURES 15 TO 30C. 



Temp., 


Density (unit = weight 
in vacuo of 1 ml. water 
at 4C.) 


Weight in grams of 1 ml. 
water, in glass container, 
in air against brass weights 


15 


0.99913 


0.99793 




16 


0.99897 


0.99780 




17 


0.99880 


0.99766 




18 


0.99862 


0.99751 




19 


0.99843 


0.99735 




20 


0.99823 


0.99718 




21 


0.99802 


0.99700 




22 


0.99780 


0.99680 




23 


0.99757 


0.99660 




24 


0.99732 


0.99638 




25 


0.99707 


0.99615 




26 


0.99681 


0.99593 




27 


0.99654 


0.99569 




28 


0.99626 


0.99544 




29 


0.99597 


0.99518 




30 


0.99567 


0.99491 





365 



366 



CALCULATIONS OF ANALYTICAL CHEMISTRY 



TABLE V. VAPOR PRESSURE OP WATER 



Temperature, 


Pressure, 


Temperature, 


Pressure, 


C. 


mm. 


C. 


mm. 





4.6 


21 


18.5 


1 


4.9 


22 


19.7 


2 


5.3 


23 


20.9 


3 


5.7 


24 


22.2 


4 


6.1 


25 


23.6 


5 


6.5 


26 


25.0 


6 


7.0 


27 


26.5 


7 


7.5 


28 


28.1 


8 


8.0 


29 


29.8 


9 


8.6 


30 


31.6 


10 


9.2 


31 


33.4 


11 


9.8 


32 


35.4 


12 


10.5 


33 


37.4 


13 


11.2 


34 


39.6 


14 


11.9 


35 


41.9 


15 


12.7 


40 


55.0 


16 


13.5 


50 


92.2 


17 


14.4 


60 


149.2 


18 


15.4 


70 


233.8 


19 


16.4 


80 


355.5 


20 


17.4 


90 


526.0 



APPENDIX 



367 



15 
TABLE VI. SPECIFIC GRAVITY OF STRONG ACIDS AT - IN VACUO 

(According to G. Lunge) 

(From Tread well and Hall' a "Analytical Chemistry," Vol. II, published by John Wiley 

& Sons, Inc., by permission) 



Specific 
gravity 

* 15 
at -^o- 

(vacuo) 


Per cent by weight 


Specific 
gravity 

* 15 
at-jr 

(vacuo) 


Per cent by weight 


HC1 


HNOa 


H 2 SO< 


HNOa 


H 2 SO 4 


1.000 


0.16 


0.10 


0.09 


1.235 


37.53 


31.70 


1.005 


1.15 


1.00 


0.95 


1.240 


38.29 


32.28 


1.010 


2.14 


1.90 


1.57 


1.245 


39.05 


32.86 


1.015 


3.12 


2.80 


2.30 


1.250 


39.82 


33.43 


1.020 


4.13 


3.70 


3.03 


1.255 


40.58 


34.00 


1.025 


5.15 


4.60 


3.76 


1.260 


41.34 


34.57 


1.030 


6.15 


5.50 


4.49 


1.265 


42.10 


35.14 


1.035 


7.15 


6.38 


5.23 


1.270 


42.87 


35.71 


1.040 


8.16 


7.26 


5.96 


1.275 


43.64 


36.29 


1.045 


9.16 


8.13 


6.67 


1.280 


44.41 


36.87 


1.050 


10.17 


8.99 


7.37 


1 . 285 


45.18 


37.45 


1.055 


11.18 


9.84 


8.07 


1.290 


45.95 


38.03 


1.060 


12.19 


10.68 


8.77 


1.295 


46.72 


38.61 


1.065 


13.19 


11.51 


9.47 


1.300 


47.49 


39.19 


1.070 


14.17 


12.33 


10.19 


1.305 


48.26 


39.77 


1.075 


15.16 


13.15 


10.90 


1.310 


49.07 


40.35 


1.080 


16.15 


13.95 


11.60 ^ 


1.315 


49.89 


40.93 


1.085 


17.13 


14.74 


12.30 


1.320 


50.71 


41.50 


1.090 


18.11 


15.53 


12.99 


1.325 


51.53 


42.08 


1.095 


19.06 


16.32 


13.67 


1.330 


52.37 


42.66 


1.100 


20.01 


17.11 


14.35 


1.335 


53.22 


43.20 


1.105 


20.97 


17.89 


15.03 


1.340 


54.07 


43.74 


1.110 


21.92 


18.67 


15.71 


1.345 


54.93 


44.28 


1.115 


22.86 


19.45 


16.36 


1.330 


55.79 


44.82 


1.120 


23.82 


20.23 


17.01 


1.355 


56.66 


45.35 


1.125 


24.78 


21.00 


17.66 


1.360 


57.57 


45.88 


1.130 


25.75 


21.77 


18.31 


1.365 


58.48 


46.41 


1.135 


26.70 


22.54 


18.96 


1.370 


59.39 


46.94 


1.140 


27.66 


23.31 


19.61 


1.375 


60.30 


47.47 


1.145 


28.61 


24.08 


20.26 


1.380 


61.27 


48.00 


1.150 


29.57 


24.84 


20.91 


1.385 


62.24 


48.53 


1.155 


30.55 


25.60 


21.55 


1.390 


63.23 


49.06 


1.160 


31.52 


26.36 


22.19 


1.395 


64.25 


49.59 


1.165 


32.49 


27.12 


22.83 


1.400 


65.30 


50.11 


1.170 


33.46 


27.88 


23.47 


.405 


66.40 


50.63 


1.175 


34.42 


28.63 


24.12 


.410 


67.50 


51.15 


1.180 


35.39 


29.38 


24.76 


.415 


68.63 


51.66 


1.185 


36.31 


30.13 


25.40 


.420 


69.80 


52.15 


1.190 


37.23 


30.88 


26.04 


.425 


70.98 


52.63 


1.195 


38.16 


31.62 


26.68 


.430 


72.17 


53.11 


1.200 


39.11 


32.36 


27.32 


.435 


73.39 


53.59 


1.205 




33.09 


27.95 


.440 


74.68 


54.07 


1.210 




33.82 


28.58 


.445 


75.98 


54.55 


1 215 




34.55 


29.21 


1.450 


77.28 


55.03 


1.220 




35.28 


29.84 


1.455 


78.60 


55.50 


1.225 




36.03 


30.48 


1.460 


79.98 


55.97 


1.230 




3 78 


ai 11 * 


1 ARK. 


01 An 


fO M*% 



368 



CALCULATIONS OF ANALYTICAL CHEMISTRY 



15 a 



SPECIFIC GRAVITY OF STRONG ACIDS AT -TO" IN VACXTO. (Continued) 
(According to G. Lunge) 



Specific 
gravity 

.15 
at-^s- 

(vacuo) 


Per cent by weight 


Specific 
gravity 

ati 5 
at ^o 

(vacuo) 


Per cent 
by 
weight 


Specific 
gravity 

.15 
at -|<r 

(vacuo) 


Per cent 
by 
weight 


HNO, 


H 2 SO 4 


H,SO 4 


H S SO 4 


1.470 
1.475 
1.480 
1.485 
1.490 
1.495 
1.500 
1.505 
1.510 
1.515 
1.520 
1.525 
1.530 
1.535 
1.540 
1.545 
1.550 
1.555 
1.560 
1.565 
1.570 
1.575 
1.580 
1.585 
1.590 
1.595 
1.600 
1.605 


82.90 
84.45 
86.05 
87.70 
89.90 
91.60 
94.09 
96.39 
98.10 
99.07 
99.67 


56.90 
57.37 
57.83 
58.28 
58.74 
59.22 
59.70 
60.18 
60.65 
61.12 
61.59 
62.06 
62.53 
63.00 
63.43 
63.85 
64.26 
64.67 
65.20 
65.65 
66.09 
66.53 
66.95 
67.40 
67.83 
68.26 
68.70 
69.13 


1.610 
1.615 
1.620 
1.625 
1.630 
1.635 
1.640 
1.645 
1.650 
1.655 
1.660 
1.665 
1.670 
1.675 
1.680 
1.685 
1.690 
1.695 
1.700 
1.705 
1.710 
1.715 
1.720 
1.725 
1.730 
1.735 
1.740 
1.745 


69.56 
70.00 
70.42 
70.85 
71.27 
71.70 
72.12 
72.55 
72.96 
73.40 
73.81 
74.24 
74.66 
75.08 
75.50 
75.94 
76.38 
76.76 
77.17 
77.60 
78.04 
78.48 
78.92 
79.36 
79.80 
80.24 
80.68 
81.12 


1.750 
1.755 
1.760 
1.765 
1.770 
1.775 
1.780 
1.785 
1.790 
1.795 
1.800 
1.805 
1.810 
1.815 
1.820 
1.825 
1.830 
1.835 
1.840 
1.8405 
1.8410 
1.8415 
1.8410 
1.8405 
1.8400 
1.8395 
1.8390 
1.8385 


81.56 
82.00 
82.44 
83.01 
83.51 
84.02 
84.50 
85.10 
85.70 
86.30 
86.92 
87.60 
88.30 
89.16 
90.05 
91.00 
92.10 
93.56 
95.60 
95.95 
96.38 
97.35 
98.20 
98.52 
98.72 
98.77 
99.12 
99.31 





































APPENDIX 



369 



TABLE VII. SPECIFIC GRAVITY OF POTASSIUM AND SODIUM HYDROXIDE 

SOLUTIONS AT 15C. 

(From Treadwell and Hall's "Analytical Chemistry," Vol. II, published by John Wiley 

<fc Sons Inc. by permission) 



Specific 
gravity 


Per cent 
KOH 


Per cent 
NaOH 

i 


Specific 
gravity 


Per cent 
KOH 


Per cent 
NaOH 


1.007 


0.9 


0.61 


1.252 


27.0 


22.64 


1.014 


1.7 


1.20 


1.263 


28.2 


23.67 


1.022 


2.6 


2.00 


1.274 


28.9 


24.81 


1.029 


3.5 


2.71 


1.285 


29.8 


25.80 


1.037 


4.5 


3.35 


1.297 


30.7 


26.83 


1.045 


5.6 


4.00 


1.308 


31.8 


27.80 


1.052 


6.4 


4.64 


1.320 


32.7 


28.83 


1.060 


7.4 


5.29 


1.332 


33.7 


29.93 


1.067 


8.2 


5.87 


1.345 


34.9 


31.22 


1.075 


9.2 


6.55 


1.357 


35.9 


32.47 


1.083 


10.1 


7.31 


1.370 


36.9 


33.69 


1.091 


10.9 


8.00 


1.383 


37.8 


34.96 


1.100 


12.0 


8.68 


1.397 


38.9 


36.25 


1.108 


12.9 


9.42 


1.410 


39.9 


37.47 


1.116 


13.8 


10.06 


1.424 


40.9 


38.80 


1.125 


14.8 


10.97 


1.438 


42.1 


39.99 


1.134 


15.7 


11.84 


1.453 


43.4 


41.41 


1.142 


16.5 


12.64 


1.468 


44.6 


42.83 


1.152 


17.6 


13.55 


1.483 


45.8 


44.38 


1.162 


18.6 


14.37 


1.498 


47.1 


46.15 


1.171 


19.5 


15.13 


1.514 


48.3 


47.60 


1.180 


20.5 


15.91 


1.530 


49.4 


49.02 


1.190 


21.4 


16.77 


1.546 


50.6 




1.200 


22 4 


17.67 


1.563 


51.9 




1.210 


23.3 


18.58 


1.580 


53.2 




1.220 


24.2 


19.58 


1.597 


54.5 




1.231 


25.1 


20.59 


1.615 


55.9 




1.241 


26.1 


21.42 


1.634 


57.5 





370 



CALCULATIONS OF ANALYTICAL CHEMISTRY 



TABLE VIII. SPECIFIC GRAVITY OF AMMONIA SOLUTIONS AT 15C. 

(According to Lunge and Wiernik) 

(From Treadwell & Hall's "Analytical Chemistry," Vol. II, published by John Wiley 

& Sons, Inc. by permission) 



Specific gravity 


Per cent NH S 


Specific gravity 


Per cent NH, 


1.000 


0.00 


0.940 


15.63 


0.998 


0.45 


0.938 


16.22 


0.996 


0.91 


0.936 


16.82 


0.994 


1.37 


0.934 


17.42 


0.992 


1.84 


0.932 


18.03 


0.990 


2.31 


0.930 


18.64 


0.988 


2.80 


0.928 


19.25 


0.986 


3.30 


0.926 


19.87 


0.984 


3.80 


0.924 


20.49 


0.982 


4.30 


0.922 


21.12 


0.980 


4.80 


0.920 


21.75 


0.978 


5.30 


0.918 


22.39 


0.976 


5.80 


0.916 


23.03 


0.974 


6.30 


0.914 


23.68 


0.972 


6.80 


0.912 


24.33 


0.970 


7.31 


0.910 


24.99 


0.968 


7.82 


0.908 


25.65 


0.966 


8.33 


0.906 


26.31 


0.964 


8.84 


0.904 


26.98 


0.962 


9.35 


0.902 


27.65 


0.960 


9.91 


0.900 


28.33 


0.958 


10.47 


0.898 


29.01 


0.956 


11.03 


0.896 


29.69 


0.954 


11.60 


0.894 


30.37 


0.952 


12.17 


0.892 


31.05 


0.950 


12.74 


0.890 


31.75 


0.948 


13.31 


0.888 


32.50 


0.946 


13.88 


0.886 


33.25 


0.944 


14.46 


0.884 


34.10 


0.942 


15.04 


0.882 


34.95 



APPENDIX 



371 



TABLE IX. IONIZATION CONSTANTS, 25C. 



Acids 





Constant 
for 1st 
hydrogen 


Constant 
for2d 
hydrogen 


Constant 
for 3d 
hydrogen 


Acetic acid, HC 2 H 3 2 


1.86 X 10-s 
5 X 10-' 
6.6 X 10~ 6 
5.5 X 10- 10 
3.3 X 10-' 
1.6 X 10-' 
8 X 10-* 
2.1 X 10- 
7.2 X 10- 10 
9.1 X 10-" 
4.0 X 10-o 
1.6 X 10- 4 
4.5 X 10~ 4 
3.8 X 10-= 
1.1 X ID" 2 
5 X 10-2 
3 ' X 10-" 
1.7 X lO- 2 
1.1 X 10~ 3 


4 X 10-* 
5 X 10-" 

1.2 X 10- 15 

4.9 X 10~ 5 
2.0 X 10~ 7 
2 X 10~ 6 
5 X 10-" 
5 X 10~ 
6.9 X 10~ 6 


6 X 10- w 
3.6 X 10~ 13 


Arsenic acid, H 3 As0 4 


Benzoic acid, HCvHjOj 


Boric acid, IIsBOj 


Carbonic acid, II 2 CO 3 


Chloracetic acid, HC 2 1L0 2 C1 


Citric acid, Ii 3 CHiO7 


Formic acid, HCH0 2 


Hydrocyanic acid, IICN 


Hydrogen sulfide, H 2 S 


Hypochlorous acid, HC10 


Lactic acid, IICjHeO) 


Nitrous acid, HNO 2 


Oxalic acid, H 2 C 2 O 4 


Phosphoric acid, H 3 PO 4 


Phosphorous acid, H 3 PO 3 


Selenious acid, H 2 Se0 3 


Sulfurous acid, H 2 SO 3 


Tartaric acid, H 2 C 4 H 4 O 6 




Bases 


Complex Ions 


Ammonium hydroxide, NH 4 OH 
Aniline, CHsNH 2 


1.75 X 10-' 
4 X lO" 10 
1.3 X 10-" 
7.4 X 10- 4 
5.6 X 10~ 4 
4.4 X 10~ 4 
2.3 X 10- 


1 
( 
( 
( 

I 
2 

t 

C 

c 

I 
I 

J 
I 
I 

A 


Lg(NH 3 ) 2 +.. 
Jd(NH,) 4 ++. 
:o(NH 3 ) 6 +++- 
:u(NH,) 4 ++. 
^i(NH,) 4 ++. 
;n(NH 3 ) 4 ++ . 
Lg(CN)r. . . 
M(CN)r... 
3u(CN)r. . . 
^e(CN), 5 ... 
Ig(CN) 4 -. . . 

(CN)r... 
I g i 4 - 


6.8 X 10-* 
2.5 X 10~ 7 
2 X 10" 34 
4.6 X 10~ 14 
5 X lO- 8 
3 X 10~ 10 
1.0 X 10~ 21 
1.4 X 10~ 17 
5.0 X 10- 2 8 
1.0 X 10-* 
4.0 X 10~ 42 
1.0 X 10- 22 
5.0 X 10- 8 i 
2.0 X 10- 55 
4.0 X 10~ 14 


Diethyl amine, (C 2 H 6 ) 2 NH 
Dimethyl amine, (CH 8 ) 2 NH. . . 
Ethyl amine, CJTsNHz 


Methyl amine, CH 3 NH 2 


Pyridine, C S H 6 N 




IgS 2 " 


^(S 2 3 )2 S . . 



372 CALCULATIONS OF ANALYTICAL CHEMISTRY 



TABLE X. SOLUBILITY PRODUCTS, Approximately 25C. 



Aluminum hydroxide, Al(OH)s 

Barium carbonate, BaCOs 

chromate, BaCrO4 

fluoride, BaF2 

iodate, Ba (163)2 

oxalate, BaC 2 04 

sulf ate, BaSO4 

Bismuth sulfide, BisSa 

Cadmium sulfide, CdS 

Calcium carbonate, CaCOa . . . 

chromate, CaCrCh . . . 

fluoride, CaF 2 

iodate, Ca(IOa)2 

oxalate, CaCsCh 

sulf ate, CaS04 

Cobalt sulfide, CoS 

Cupric sulfide, CuS 

Cuprous chloride, CuCl 

bromide, CuBr 

iodide, Cul 

sulfide, Cu2S 

thiocyanate,CuCNS. 

Ferric hydroxide, Fe(OH)s 

Ferrous hydroxide, Fe(OH)a . . 
sulfide, FeS 

Lead carbonate, PbCOa 

chloride, PbCh 

chromate, PbCrO4 

fluoride, PbF 2 

iodate, Pb(I0 3 )2 

iodide, Pbl? 

oxalate, PbC2O4 

phosphate, Pb3(PO 4 )2 . . . 

sulf ate, PbS04 

sulfide, PbS 



3.7 X 
8.1 X 1(T* 
3.0 X 10- w 
1.7 X 10 

6.0 X 10-1 
1.7 X 10^ 

1.1 X 10- 10 
1.6 X 10~ 
3.6 X 10' 
1.6 X 10-* 

2.3 X 

3.2 X 

6.4 X 10-* 
2.6 X 10- 

6.4 X 10-' 
3.0 X 10-* 
8.0 X 10- 

1.0 X 10- 

4.1 X 10-* 
5,0 X 10 l2 

1.0 X 10-<* 
1.6 X 10 

1.1 X 10- 

1.6 X 10-M 

1.5 X 



5.6 X 

2.4 X 10-* 

1.8 X 10~" 

3.7 X 10-8 

9.8 X 10-" 

2.4 X 10- 
3.3 X 

1.5 X 

1.1 X lO-* 

4.2 X 10* 28 



Magnesium carbonate, MgCOj. . 
fluoride, MgF2 ..... 
hydroxide, Mg(OH)j 
oxalate, 



Manganese hydroxide, Mn(OH>2 
sulflde, MnS ....... 

Mercurous chloride, HgsCh ..... 
bromide, Hg2Bra ..... 
iodide, Hg 2 I ........ 



Nickel sulfide, NiS . 



Silver bromate, AgBrOa 

bromide, AgBr 

carbonate, Ag2COs 

chloride, AgCl 

chromate, Ag2CrO4 

cyanide, Ag2(CN>2 

hydroxide, AgOH 

iodate, AglOa 

iodide, Agl 

nitrite, AgNO2 

oxalate, Ag2C2(>4 

phosphate, AgsPO4 

sulf ate, Ag2SO4 

sulfide, A gsS 

thiocyanate, AgCNS 

Strontium carbonate, SrCOs .... 
chromate, SrCrO* .... 

fluoride, SrF 3 

oxalate, SrC 2 04 

sulf ate, SrSO4 

Zinc carbonate, ZnCOa 

hydroxide, Zn(OH) 2 

sulfide, ZnS 



2.6 X 10~* 
8.4 X 10-* 
3.4 X 10-" 
8.6 X 10-* 

4.0 X 
1.4 X 

1.1 X 
1.4 X 

1.2 X 

1.4 X 10-* 



5.0 
5.0 
6.2 
1.0 
9.0 
1.2 
1.5 
2.0 
1.0 
7.0 
1.3 
1.8 
7.0 
1.6 
1.0 



xio- 



X 10-12 
X 10-w 
X 10-u 
X lO-" 
X10- 

xio- 

XlO-w 
X 10-* 
X 10-" 
X 10- 

xio- 
x io-< 

X 10-n 



1.6 X 10- 
3.0 X 10-* 
2.8 X 10- 
5.6 X 10~ 
2.8 X 10-7 

3.0 X 10-* 
1.8 X 10-" 
1.2 X 



APPENDIX 



373 



TABLE XI. SPECIFIC OXIDATION-REDUCTION POTENTIALS 

[Temperature = 25C. Solution concentrations are 1 molar unless other- 
wise specified. Gases (g) are at 1 atmosphere pressure] 



Half-cell Reaction 
K * K+ + 
Sr <=* Sr ++ + 2e 



Ca <= Ca- 1 " 1 - + 2 



Mg <= Mg++ + 26 



Mn 3F* Mn++ + 2e 

Zn ^ Zn ++ + 2c 

Cr ? Cr+++ + 3e 

S~ <=* S + 2e 

H 2 C 2 4 (aq) <=* 2C0 2 (g) + 2H+ + 2e 

Fe ?= Fe ++ + 2e 

H 2 (g) ?=t 2H+ (10-* M) + 2e 

Cd ?=* Cd ++ + 26 

Co *= Co ++ + 2c 

Ni ? Ni ++ + 2 

Sn ? Sn- 1 -*- + 2e 

Pb ?= Pb++ + 2e 

Fe -Ft Fe +++ + 3e * 

H 2 (g) *^ 2H+ + 2c 

Sn ++ ?i 8n++++ + 2e 

H 2 S ^ S + 2H+ + 2c 

Bi + 4C1- & BiCLr + 3e 

Sb + H 2 O 3=4 SbO + + 2H + + 3e 

Ag + 01- SF AgCl + e 

As + 3H 2 O ?=t H^O 8 + 3H+ + 3e 

2Hg + 2C1- ?= HgjCls + 2e (calomel cell) 

Bi + H 2 O *=* BiO + + 2H+ + 3e 

Cu ?= Cu ++ + 2e 

Fe(CN),= ?=t Fe(CN),- + e 

HaAsO, + H 2 O <=* H^sO 4 + 2H+ + 2e 

21- ^ I 2 + 2e 

MnO 4 ~ *^ MnO 4 - + 6 

H 2 2 ?=t 2 (g) + 2H+ + 26 

C,H 4 (OH) 2 * C.H.Os + 2H+ + 2e (quinhydrone) 

MnO 2 + 4OH- * MnO 4 - + 2H 2 O + 2 

Fe ++ <= Fe- 1 "*" 1 - + 

Ag ;F Ag + + 6 

2H 2 O ?=t O 2 (g) + 4H+ (10- 7 M) + 4e 

Hg ?=t Hg++ + 2e 



E 

-2.992 
-2.92 
-2.90 
-2.87 
-2.713 
-2.40 
-1.67 
- 1.05 
-0.758 
-0.71 
-0.51 
-0.49 
-0.441 
-0.414 
-0.398 
-0.277 
-0.22 
-0.136 
-0.122 
-0.045 

0.000 
+0.13 
+0.141 
+0.168 
+0.212 
+0.222 
+0.24 
+0.285 
+0.32 
+0.344 
+0.40 
+0.49 
+0.535 
+0.66 
+0.68 
+0.700 
+0.71 
+0.747 
+0.799 
+0.815 
+0.86 



374 CALCULATIONS OF ANALYTICAL CHEMISTRY 
SPECIFIC OXIDATION-REDUCTION POTENTIALS. (Continued) 
Half-cell Reaction E 

NO + 2H 2 ^ NO," + 4H+ + 3e +0.94 

HN0 2 + H 4 ^ NO,' + 3H+ + 2e +0.96 

2Br-^Br 2 (aq)+2<- +J-065 

2Cr +^ + 7H 2 O^Cr 2 7 - + 14H + + 6t +1-30 

Mn ++ + 2H 2 ^ Mn0 2 + 4H + + 2c +1-33 



Ce +++ ^ 

Mn ++ + 4H 2 ^ MnOr + 8H + + 5 +1-52 

Mn0 2 + 2H 2 & MnOr + 4H+ + 3e +1-63 

PbS0 4 + 2H 2 ^ Pb0 2 + 4H + + SOr + 2c +1-70 



APPENDIX 



375 



TABLE XII. FORMULA WEIGHTS 

(These weights cover most of the compounds encountered in the problems of 

this text) 



462.35 

AgBr .................... 187.80 

AgBrOs .................. 235.80 

AgCl .................... 143.34 

Agl ...................... 234.80 

AgNO, ................... 169.89 

Ag 3 PO 4 ................... 418.65 

Ag 2 SO 4 ................... 311.82 

AlBr, .................... 266.72 

A1 2 O, .................... 101.94 

Al(OH), .................. 77.99 

A1 2 (SO 4 ) S ................. 342.12 

A1 2 (SO 4 ),.18H 2 O ........... 666.4 1 



197.82 

As 2 O 6 .................... 229.H2 

AsjS, ..................... 2-16.00 

Ba 3 (AsO 4 ) 2 ................ 689.90 

BaBr 2 .................... 297.19 

BaCl 2 .................... 208.27 

BaCl 2 .2II 2 ............... 244.31 

BaCO, ................... 197.37 

BaQA .................. 225.38 

BaF 2 ..................... 175.36 

BaI 2 ..................... 391.20 

Ba(IO 3 ) 2 ................. 487.20 

BaO ..................... 153.30 

Ba(OH) 2 ................. 171.38 

Ba(OH) 2 .8H 2 ............ 315.50 

BaSO 4 ................... 233.42 



BeO. 



25.02 



Bi(NO 3 ) 3 .5H 2 O 485.10 

BiO 2 241.00 

BizOs 466.00 

BiOHCO, 286.02 

Bi 2 S 3 514.18 

CaCl 2 110.99 

CaCO, 100.09 

CaF 2 78.08 

Ca(NO,) 2 164.10 

CaO 56.08 

Ca(OH) 2 74.10 

Ca 3 (PO 4 ) 2 310.20 

3Ca 3 (PO 4 ) 2 .CaCl 2 1041.59 

CaSO 4 136.11 



CeO 2 

Ce(SO 4 ) 2 .2(NH 4 ) 2 SO 4 .2H 2 O 

CH,COOH (acetic acid) . . . 

(CH,CO) 2 O 

CeHsCOOH (benzoic acid). 

CO 2 

CO(NH 2 ) 2 

CS(NH 2 ) 2 



CrCl, 

Cr 2 O 3 

Cr 2 (SO 4 ), 

CuO 

Cu 2 (OH) 2 CO, 

CuS 

Cu 2 S 

CuSO 4 .5H 2 O 



Fed, 

FeCl 3 .6H 2 

FeCO, 

Fe(CrO 2 ) 2 

Fe(N0 3 ) 3 .6H 2 O 

F6O 

Fe 2 0, 

Fe 3 O 4 

Fe(OH), 

FeS 2 

Fe 2 Si 

FeSO 4 .7H 2 O 

Fe 2 (S0 4 ) 3 

Fe 2 (SO 4 ),.9II 2 O 

FeSO 4 . (NH 4 ) 2 SO 4 .6H 2 O . 



HBr 

HCH0 2 (formic acid) 

HC 2 HsO 2 (acetic acid) 

HCyHtOi (benzoic acid) . . . 

HC1 

HC1O 4 

H 2 C 2 O 4 .2H 2 O (oxalic acid) . 

HCOOH (formic acid) 

IINO 3 

H 2 O 

H 2 2 

H,P0 3 

H 8 P0 4 

HjS 

H 2 SO, 

H 2 SO 4 



172.13 
632.56 

60.05 

102.09 

122.12 

44.01 

60.06 

76.12 

158.38 
152.02 
389.20 

79.57 
221.17 

95.63 
159.20 
249.71 

162.22 
270.32 
115.86 
223.87 
349.97 
71.84 
159.70 
231.55 
106.87 
119.97 
139.76 
278.02 
399.88 
562.02 
392.15 

80.92 
46.03 
60.05 
122.12 
36.47 
100.46 
126.07 
46.03 
63.02 
18.02 
34.02 
82.00 
98.00 
34.08 
82.06 
98.08 



376 



CALCULATIONS OF ANALYTICAL CHEMISTRY 



FORMULA WEIGHTS. (Continued) 



HgjBrj 561.06 

Hg 2 Cl 2 472.14 

Hg 2 I 2 655.06 

KA1(SO 4 ) 2 .12H 2 O 474.38 

K,AsO 4 256.20 

KBrO, 167.01 

KC1 74.56 

KC1O, 122.56 

KC1O 4 138.55 

KCN 65.11 

KCNS 97.17 

K 2 CO, 138.20 

K 2 CrO 4 194.20 

K 2 Cr 2 O 7 294.21 

KsFe(CN), 329.25 

K,Fe(CN),,.3H 2 O 422.39 

KHCJ^O, (tartrate) 188.18 

KHC 8 H4O 4 (phthalate) 204.16 

KHCOa 100.11 

KHCjO 4 128.12 

KHCzCVHzO 146.14 

KHC 2 O 4 .H 2 C 2 O 4 .2H 2 254.19 

KH(IO,) 2 389.94 

KHSO 4 136.16 

KI 166.02 

KIO, 214.02 

KMnO 4 158.03 

KNaC 4 H 4 O,.4H 2 282.23 

KNaCO, 122.11 

KNO 2 85.10 

KNO 101.10 

K 2 94.19 

KOH 56.10 

K,PO 4 212.27 

KjPtCU 486.16 

K 2 SO 4 174.25 

K 2 SO 4 . A1 2 (S0 4 ),.24H 2 948.76 

K 2 SO 4 .Cr 2 (SO 4 ) 3 .24H 2 998.84 

LiCl 42.40 

LiCOa 73.89 

Li 2 29.88 

LiOH 23.95 

MgCl 2 95.23 

MgCO, 84.33 

MgNH 4 AsO 4 181.27 

MgNH 4 PO 4 137.33 

MgO 40.32 

Mg(OH) 2 58.34 

MgzPjO? 222.60 

MgSO 4 120.38 

MgSO 4 .7H 2 246.49 



Mn 2 O, ................... 157.86 

Mn,O 4 ................... 228.79 

Mn 2 P 2 O7 ................. 283.82 

MoO, .................... 143.95 

Mo^Ow .................. 2894.80 

MoS, .................... 192.13 

NaaAsOa ................. 191.91 

201.27 

........... 381.43 

NaBr .................... 102.91 

NaBrOa .................. 150.91 

NaCHO 2 (formate) ........ 68.01 

NaC 2 H s O 2 (acetate) ........ 82.04 

NaCl .................... 58.45 

NaCN ................... 49.02 

Na-iCOa .................. 106.00 

NazC 2 O 4 .................. 134.01 

NaaHAsOa ................ 169.91 

NaHCO, ................. 84.01 

NaHC 2 O 4 ................. 112.03 

NazHPO 4 ................. 141.98 

Na 2 HPO 4 .12H 2 .......... 258.17 

NaHS .................... 56.07 

NaH 2 PO 4 ................. 119.99 

NaH 2 PO 4 .H 2 ............ 138.01 

Nal ..................... 149.92 

NaKCOa ................. 122.11 

NaNO 2 ................... 69.01 

NaNOa ................... 85.01 

Na-iO .................... 61.99 

NaA .................... 77.99 

NaOH ................... 40.00 

Na s PO 4 .................. 163.97 

Na 8 PO 4 .12H 2 ............ 358.17 

Na^S ..................... 78.05 

126.05 

............ 322.21 

158.11 
248.19 



MnO. 
MnO 2 . 



70.93 
86.93 



NH, ..................... 17.03 

NH 4 C1 ................... 53.50 

(NH 4 ) 2 C 2 O 4 .H 2 ........... 142.12 

(NH4) 2 HP0 4 .............. 132.07 

NH 4 OH .................. 35.05 

(NH 4 ),PO 4 .12MoO 3 ........ 1876.53 

(NH 4 ) 2 PtCl, .............. 444.05 

(NH 4 ) 2 SO 4 ................ 132.14 

NO ...................... 30.01 

NO 2 ..................... 46.01 

N 2 O, ..................... 76.02 

PbCl 2 .................... 278.12 

PbCIF ................... 261.67 



APPENDIX 



377 



FORMULA WEIGHTS. (Continued) 



PbC 2 O 4 ................... 295.23 

PbCrO 4 .................. 323.22 

PbI 2 ..................... 461.05 

Pb(IO,) 2 ................. 557.05 

Pb(NO,) 2 ................. 331.23 

PbO ..................... 223.21 

PbO 2 .................... 239.21 

Pb 2 O, .................... 462.42 

Pb,O 4 .................... 685.63 

Pb,(PO 4 )2 ................ 811.59 

PbS0 4 ................... 303.27 

PdI 2 ..................... 360.54 

P.O S ..................... 141.96 



291.52 

Sb 2 O 4 .................... 307.52 

Sb 2 O 6 .................... 323.52 

Sb 2 S 8 .................... 339.70 

SiCU ..................... 169.89 

SiF 4 ..................... 104.06 

Si0 2 ..................... 60.06 



SnCl 2 189.61 

SnCU 260.53 

SnO 2 150.70 

SO 2 64.06 

SO, 80.06 

SrCl 2 .6H 2 266.64 

SrCO, 147.64 

SrO 103.63 



TiO 2 



79.90 



UO S ..................... 286.14 

U 3 O 8 ..................... 842.21 

WO, ..................... 231.92 



178.40 
ZnO ..................... 81.38 

Zn 2 P 2 7 .................. 304.72 

ZnSO 4 .7H 2 .............. 161.44 

ZrO 2 ..................... 123.22 



378 CALCULATIONS OF ANALYTICAL CHEMISTRY 

LOGARITHMS OF NUMBERS 



Natural 
numbers 





1 


2 


3 


4 


5 


6 


7 


8 


9 


Proportional parts 


1 


2 


3 


4 


5 


6 


7 


8 


9 


10 


0000 


0043 


0086 


0128 


0170 


0212 


0253 


0294 


0334 


0374 


4 


8 


12 


17 


21 


25 


29 


33 


37 


11 


0414 


0453 


0492 


0531 


0569 


0607 


0645 


0682 


0719 


0755 


4 


8 


11 


15 


19 


23 


26 


30 


34 


12 


0792 


0828 


0864 


0899 


0934 


0969 


1004 


1038 


1072 


1106 


3 


7 


10 


14 


17 


21 


24 


28 


31 


13 


1139 


1173 


1206 


1239 


1271 


1303 


1335 


1367 


1399 


1430 


3 


6 


10 


13 


16 


19 


23 


26 


29 


14r 


1461 


1492 


1523 


1553 


1584 


1614 


1644 


1673 


1703 


1732 


3 


6 


9 


12 


15 


18 


21 


24 


27 


15 


1761 


1790 


1818 


1847 


1875 


1903 


1931 


1959 


1987 


2014 


3 


6 


8 


11 


14 


17 


20 


22 


25 


16 


2041 


2068 


2095 


2122 


2148 


2175 


2201 


2227 


2253 


2279 


3 


5 


8 


11 


13 


16 


18 


21 


24 


17 


2304 


2330 


2355 


2380 


2405 


2430 


2455 


2480 


2504 


2529 


2 


5 


7 


10 


12 


15 


17 


20 


22 


18 


2553 


2577 


2601 


2625 


2648 


2672 


2695 


2718 


2742 


2765 


2 


5 


7 


9 


12 


14 


16 


19 


21 


19 


2788 


2810 


2833 


2856 


2878 


2900 


2923 


2945 


2967 


2989 


2 


4 


7 


9 


11 


13 


16 


18 


20 


20 


3010 


3032 


3054 


3075 


3096 


3118 


3139 


3160 


3181 


3201 


2 


4 


6 


8 


11 


13 


15 


17 


19 


21 


3222 


3243 


3263 


3284 


3304 


3324 


3345 


3365 


3385 


3404 


2 


4 


6 


8 


10 


12 


14 


16 


18 


22 


3424 


3444 


3464 


3483 


3502 


3522 


3541 


3560 


3579 


3598 


2 


4 


6 


8 


10 


12 


14 


15 


17 


23 


3617 


3636 


3655 


3674 


3692 


3711 


3729 


3747 


3766 


3784 


2 


4 


6 


7 


9 


11 


13 


15 


17 


24 


3802 


3820 


3838 


3856 


3874 


3892 


3909 


3927 


3945 


3962 


2 


4 


5 


7 


9 


11 


12 


14 


16 


25 


3979 


3997 


4014 


4031 


4048 


4065 


4082 


4099 


4116 


4133 


2 


3 


5 


7 


9 


10 


12 


14 


15 


26 


4150 


4166 


4183 


4200 


4216 


4232 


4249 


4265 


4281 


4298 


2 


3 


5 


7 


8 


10 


11 


13 


15 


27 


4314 


4330 


4346 


4362 


4378 


4393 


4409 


4425 


4440 


4456 


2 


3 


5 


6 


8 


9 


11 


13 


14 


28 


4472 


4487 


4502 


4518 


4533 


4548 


4564 


4579 


4594 


4609 


2 


3 


5 


6 


8 


9 


11 


12 


14 


29 


4624 


4639 


4654 


4669 


4683 


4698 


4713 


4728 


4742 


4757 


1 


3 


4 


6 


7 


9 


10 


12 


13 


30 


4771 


4786 


4800 


4814 


4829 


4843 


4857 


4871 


4886 


4900 


1 


3 


4 


6 


7 


9 


10 


11 


13 


31 


4914 


4928 


4942 


4955 


4969 


4983 


4997 


5011 


5024 


5038 


1 


3 


4 


6 


7 


8 


10 


11 


12 


32 


5051 


5065 


5079 


5092 


5105 


5119 


5132 


5145 


5159 


5172 


1 


3 


4 


F 


7 


8 


9 


11 


12 


33 


5185 


5198 


5211 


5224 


5237 


5250 


5263 


5270 


5289 


5302 


1 


3 


4 


5 


6 


8 


9 


10 


12 


34 


5315 


5328 


5340 


5353 


5366 


5378 


5391 


5403 


5416 


5428 


1 


3 


4 


p 


6 


8 


9 


10 


11 


35 


5441 


5453 


5465 


5478 


5490 


5502 


5514 


5527 


5539 


5551 


1 


2 


4 


5 


6 


7 


9 


10 


11 


36 


5563 


5575 


5587 


5599 


5611 


5623 


5635 


5647 


5658 


5670 


1 


2 


4 


5 


6 


7 


8 


10 


11 


37 


5682 


5694 


5705 


5717 


5729 


5740 


5752 


5763 


5775 


5786 


1 


2 


3 


5 


6 


7 


8 


9 


10 


38 


5798 


5809 


5821 


5832 


5843 


5855 


5866 


5877 


5888 


5899 


1 


2 


3 


5 


6 


7 


8 


9 


10 


39 


5911 


5922 


5933 


5944 


5955 


5966 


5977 


5988 


5999 


6010 


1 


2 


3 


4 


5 


7 


8 


9 


10 


40 


6021 


6031 


6042 


6053 


6064 


6075 


6085 


6096 


6107 


6117 


1 


2 


3 


4 


c 


6 


8 


9 


10 


41 


6128 


6138 


6149 


6160 


6170 


6180 


6191 


6201 


6212 


6222 


1 


2 


3 


4 


5 


6 


7 


8 


9 


42 


6232 


6243 


6253 


6263 


6274 


6284 


6294 


6304 


6314 


6325 


1 


2 


3 


4 


5 


6 


7 


8 


9 


43 


6335 


6345 


6355 


6365 


6375 


6385 


6395 


6405 


6415 


6425 


1 


2 


3 




5 


6 


7 


8 


9 


44 


6435 


6444 


6454 


6464 


6474 


6484 


6493 


6503 


6513 


6522 


1 


2 


2 




5 


6 


7 


8 


9 


45 


6532 


6542 


6551 


6561 


6571 


6580 


6590 


6599 


6609 


6618 


1 


2 


3 




5 


6 


7 


8 


9 


46 


6628 


6637 


6646 


6656 


6665 


6675 


6684 


6693 


6702 


6712 


1 


2 


2 




5 


6 


7 


7 


8 


47 


6721 


6730 


6739 


6749 


6758 


6767 


6776 


6785 


6794 


6803 


1 


2 


2 




c 


5 


6 


7 


8 


48 


6812 


6821 


6830 


6839 


6848 


6857 


6866 


6875 


6884 


6893 


1 


2 


2 




4 


5 


6 


7 


8 


49 


6902 


6911 


6920 


6928 


6937 


6946 


6955 


6964 


6972 


6981 


1 


2 


2 




4 


e 


6 


7 


8 


50 


6990 


6998 


7007 


7016 


7024 


7033 


7042 


7050 


7059 


7067 


1 


2 




3 


4 


5 


6 


7 


8 


51 


7076 


7084 


7093 


7101 


7110 


7118 


7126 


7135 


7143 


7152 


1 


2 




3 


4 


5 


6 


7 


8 


52 


7160 


7168 


7177 


7185 


7193 


7202 


7210 


7218 


7226 


7235 


1 


2 




t 


4 


5 


6 


7 


7 


53 


7243 


7251 


7259 


7267 


7275 


7284 


7292 


7300 


7308 


7316 


1 


2 







4 


c 


6 


6 


7 


54 


7324 


7332 


7340 


7348 


7356 


7364 


7372 


7380 


7388 


7396 


1 


2 




I 


4 


5 


6 


6 


7 



APPENDIX 
LOGARITHMS. (Continued) 



379 



Natural 
numbers 





1 


2 


3 


4 


5 


6 


7 


8 


9 


Proportional parts 


1 


2 


3 


4 


5 


6 


7 


8 


9 


55 


7404 


7412 


7419 


7427 


7435 


7443 


7451 


7459 


7466 


7474 


1 


2 


2 


3 


4 


5 


5 


6 


7 


56 


7482 


7490 


7497 


7505 


7513 


7520 


7528 


7536 


7543 


7551 


1 


2 


2 


3 


4 


5 


5 


6 


7 


57 


7559 


7566 


7574 


7582 


7589 


7597 


7604 


7612 


7619 


7627 


1 


2 


2 


3 


4 


5 


5 


6 


7 


58 


7634 


7642 


7649 


7657 


7664 


7672 


7679 


7686 


7694 


7701 


1 


1 


2 


3 


4 


4 


5 


6 


7 


59 


7709 


7716 


7723 


7731 


7738 


7745 


7752 


7760 


7767 


7774 


1 


1 


2 


3 


4 


4 


5 


6 


7 


60 


7782 


7789 


7796 


7803 


7810 


7818 


7825 


7832 


7839 


7846 


1 


1 


2 


3 


4 


4 


5 


6 


6 


61 


7853 


7860 


7868 


7875 


7882 


7889 


7896 


7903 


7910 


7917 


1 


1 


2 


3 


4 


4 


5 


6 


6 


62 


7924 


7931 


7938 


7945 


7952 


7959 


7966 


7973 


7980 


7987 


1 


1 


2 


3 


3 


4 


5 


6 


6 


63 


7993 


8000 


8007 


8014 


8021 


8028 


8035 


8041 


8048 


8055 


1 


1 


2 


3 


3 


4 


5 


5 


6 


64 


8062 


8069 


8075 


8082 


8089 


8096 


8102 


8109 


8116 


8122 


1 


1 


2 


3 


3 


4 


5 


5 


6 


65 


8129 


8136 


8142 


8149 


8156 


8162 


8169 


8176 


8182 


8189 


1 


1 


2 


3 


3 


4 


5 


5 


6 


66 


8195 


8202 


8209 


8215 


8222 


8228 


8235 


8241 


8248 


8254 


1 


1 


2 


3 


3 


4 


5 


5 


6 


67 


8261 


8267 


8274 


8280 


8287 


8293 


8299 


8306 


8312 


8319 


1 


1 


2 


3 


3 


4 


5 


5 


6 


68 


8325 


8331 


8338 


8344 


8351 


8357 


8363 


8370 


8376 


8382 


1 


1 


2 


3 


3 


4 


4 


5 


6 


69 


8388 


8395 


8401 


8407 


8414 


8420 


8426 


8432 


8439 


8445 


1 


1 


2 


2 


3 


4 


4 


5 


6 


70 


8451 


8457 


8463 


8470 


8476 


8482 


8488 


8494 


8500 


8506 


1 


1 


2 


2 


3 


4 


4 


5 


6 


71 


8513 


8519 


8525 


8531 


8537 


8543 


8549 


8555 


8561 


8567 


1 


1 


2 


2 


3 


4 


4 


5 


5 


72 


8573 


8579 


8585 


8591 


8597 


8603 


8609 


8615 


8621 


8627 


1 


1 


2 


2 


3 


4 


4 


5 


5 


73 


8633 


8639 


8645 


8651 


8657 


8663 


8669 


8675 


8681 


8686 


1 


1 


2 


2 


3 


4 


4 


5 


5 


74 


8692 


8698 


8704 


8710 


8716 


8722 


8727 


8733 


8739 


8745 


1 


1 


2 


2 


3 


4 


4 


5 


5 


75 


8751 


8756 


8762 


8768 


8774 


8779 


8785 


8791 


8797 


8802 


1 


1 


2 


2 


3 


3 


4 


5 


5 


76 


8808 


8814 


8820 


8825 


8831 


8837 


8S42 


8848 


8854 


8859 


1 


1 


2 


2 


3 


3 


4 


5 


5 


77 


8865 


8871 


8876 


8882 


8887 


8893 


8899 


8904 


8910 


8915 


1 


1 


2 


2 


3 


3 


4 


4 


5 


78 


8921 


8927 


8932 


8938 


8943 


8949 


8954 


8960 


8965 


8971 


1 


1 


2 


2 


3 


3 


4 


4 


5 


79 


8976 


8982 


8987 


8993 


8998 


9004 


9009 


9015 


9020 


9026 


1 


1 


2 


2 


3 


3 


4 


4 


5 


80 


9031 


9036 


9042 


9047 


9053 


9058 


9063 


9&69 


9074 


9079 


1 


1 


2, 


2 


3 


3 


4 


4 


5 


81 


9085 


9090 


9096 


9101 


9106 


9112 


9117 


9122 


9128 


9133 


1 


1 


2 


2 


3 


3 


4 


4 


5 


82 


9138 


9143 


9149 


9154 


9159 


9165 


9170 


9175 


9180 


9186 


1 


1 


2 


2 


3 


3 


4 


4 


5 


83 


9191 


9196 


9201 


9206 


9212 


9217 


9222 


9227 


9232 


9238 


1 


1 


2 


2 


3 


3 


4 


4 


5 


84 


9243 


9248 


9253 


9258 


9263 


9269 


9274 


9279 


9284 


9289 


1 


1 


2 


2 


3 


3 


4 


4 


5 


85 


9294 


9299 


9304 


9309 


9315 


9320 


9325 


9330 


9335 


9340 


1 


1 


2 


2 


3 


3 


4 


4 


5 


86 


9345 


9350 


9355 


9360 


9365 


9370 


9375 


9380 


9385 


9390 


1 


1 


2 


2 


3 


3 


4 


4 


5 


87 


9395 


9400 


9405 


9410 


9415 


9420 


9425 


9430 


9435 


9440 





1 


1 


2 


2 


3 


3 


4 


4 


88 


9445 


9450 


9455 


9460 


9465 


9469 


9474 


9479 


9484 


9489 





1 


1 


2 


2 


3 


3 


4 


4 


89 


9494 


9499 


9504 


9509 


9513 


9518 


9523 


9528 


9533 


9538 





1 


1 


2 


2 


3 


3 


4 


4 


90 


9542 


9547 


9552 


9557 


9562 


9566 


9571 


9576 


9581 


9586 





1 


1 


2 


2 


3 


3 


4 


4 


91 


9590 


9595 


9600 


9605 


9609 


9614 


9619 


9624 


9628 


9633 





1 


1 


2 


2 


3 


3 


4 


4 


92 


9638 


9643 


9647 


9652 


9657 


9661 


9666 


9671 


9675 


9680 





1 


1 


2 


2 


3 


3 


4 


4 


93 


9685 


9689 


9694 


9699 


9703 


9708 


9713 


9717 


9722 


9727 





1 


1 


2 


2 


3 


3 


4 


4 


94 


9731 


9736 


9741 


9745 


9750 


9754 


9759 


9763 


9768 


9773 





1 


1 


2 


2 


3 


3 


4 


4 


95 


9777 


9782 


9786 


9791 


9795 


9800 


9805 


9809 


9814 


9818 





1 


1 


2 


2 


3 


3 


4 


4 


96 


9823 


9827 


9832 


9836 


9841 


9845 


9850 


9854 


9859 


9863 





1 


1 


2 


2 


3 


3 


4 


4 


97 


9868 


9872 


9877 


9881 


9886 


9890 


9894 


9899 


9903 


9908 





1 


1 


2 


2 


3 


3 


4 


4 


98 


9912 


9917 


9921 


9926 


9930 


9934 


9939 


9943 


9948 


9952 





1 


1 


2 


2 


3 


3 


4 


4 


99 


9956 


9961 


9965 


9969 


9974 


9978 


9983 


9987 


9991 


9996 





1 


1 


2 


2 


3 


3 


3 


4 



380 CALCULATIONS OF ANALYTICAL CHEMISTRY 

ANTILOGABITHMS 



Logarithms 





1 


2 


3 


4 


5 


G 


7 


8 


9 


Proportional parts 


1 


2 


3 


4 


5 


6 


7 


8 


9 


.00 


1000 


1002 


1005 


1007 


1009 


1012 


1014 


1016 


1019 


1021 








1 


1 


1 


1 


2 


2 


2 


.01 


1023 


1026 


1028 


1030 


1033 


1035 


1038 


1040 


1042 


1045 








1 


1 


1 


1 


2 


2 


2 


.02 - 


1047 


1050 


1052 


1054 


1057 


1059 


1062 


1064 


1067 


1069 








1 


1 


1 


1 


2 


2 


2 


.03 


1072 


1074 


1076" 


1079 


1081 


1084 


1086 


1089 


1091 


1094 








1 


1 


1 


1 


2 


2 


2 


.04 


1096 


1099 


1102 


1104 


1107 


1109 


1112 


1114 


1117 


1119 





1 


1 


1 


1 


2 


2 


2 


2 


.05 


1122 


1125 


1127 


1130 


1132 


1135 


1138 


1140 


1143 


1146 





1 


1 


1 


1 


2 


2 


2 


2 


.06 


1148 


1151 


1153 


1156 


1159 


1161 


1164 


1167 


1169 


1172 





1 


1 


1 


1 


2 


2 


2 


2 


.07 


1175 


1178 


1180 


1183 


1186 


1189 


1191 


1194 


1197 


1199 





1 


1 


1 


1 


2 


2 


2 


2 


.08 


1202 


1205 


1208 


1211 


1213 


1216 


1219 


1222 


1225 


1227 





1 


1 


1 


1 


2 


2 


2 


3 


.09 


1230 


1233 


1236 


1239 


1242 


1245 


1247 


1250 


1253 


1256 





1 


1 


1 


1 


2 


2 


2 


3 


.10 


1259 


1262 


1265 


1268 


1271 


1274 


1276 


1279 


1282 


1285 





1 


1 


1 


1 


2 


2 


2 


3 


.11 


1288 


1291 


1294 


1297 


1300 


1303 


1306 


1309 


1312 


1315 





1 


1 


1 


2 


2 


2 


2 


3 


.12 


1318 


1321 


1324 


1327 


1330 


1334 


1337 


1340 


1343 


1346 





1 


1 


1 


2 


2 


2 


2 


3 


.13 


1349 


1352 


1355 


1358 


1361 


1365 


1368 


1371 


1374 


1377 





1 


1 


1 


2 


2 


2 


3 


3 


.14 


1380 


1384 


1387 


1390 


1393 


1396 


1400 


1403 


1406 


1409 





1 


1 


1 


2 


2 


2 


3 


3 


.15 


1413 


1416 


1419 


1422 


1426 


1429 


1432 


1435 


1439 


1442 





1 


1 


1 


2 


2 


2 


3 


3 


.16 


1445 


1449 


1452 


1455 


1459 


1462 


1466 


1469 


1472 


1476 





1 


1 


1 


2 


2 


2 


3 


3 


.17 


1479 


1483 


1486 


1489 


1493 


1496 


1500 


1503 


1507 


1510 





1 


1 


1 


2 


2 


2 


3 


3 


.18 


1514 


1517 


1521 


1524 


1528 


1531 


1535 


1538 


1542 


1545 





1 


1 


1 


2 


2 


2 


3 


3 


.19 


1549 


1552 


1556 


1560 


1563 


1567 


1570 


1574 


1578 


1581 





1 


1 


1 


2 


2 


3 


3 


3 


.20 


1585 


1589 


1592 


1596 


1600 


1603 


1607 


1611 


1614 


1618 





1 


1 


1 


2 


2 


3 


3 


3 


.21 


1622 


1626 


1629 


1633 


1637 


1641 


1644 


1648 


1652 


1656 





1 


1 


2 


2 


2 


3 


3 


3 


.22 


1660 


1663 


1667 


1671 


1675 


1679 


1683 


1687 


1690 


1694 





1 


1 


2 


2 


2 


3 


3 


3 


.23 


1698 


1702 


1706 


1710 


1714 


1718 


1722 


1726 


1730 


1734 





1 


1 


2 


2 


2 


3 


3 


4 


.24 


1738 


1742 


1746 


1750 


1754 


1758 


1762 


1766 


1770 


1774 





1 


1 


2 


2 


2 


3 


3 


4 


.25 


1778 


1782 


1786 


1791 


1795 


1799 


1803 


1807 


1811 


1816 





1 


1 


2 


2 


2 


3 


3 


4 


.26 


1820 


1824 


1828 


1832 


1837 


1841 


1845 


1849 


1854 


1858 





1 


1 


2 


2 


3 


3 


3 


4 


.27 


1862 


1866 


1871 


1875 


1879 


1884 


1888 


1892 


1897 


1901 





1 


1 


2 


2 


3 


3 


3 


4 


.28 


1905 


1910 


1914 


1919 


1923 


1928 


1932 


1936 


1941 


1945 





1 


1 


2 


2 


3 


3 


4 


4 


.29 


1950 


1954 


1959 


1963 


1968 


1972 


1977 


1982 


1986 


1991 





1 


1 


2 


2 


3 


3 


4 


4 


.30 


1995 


2000 


2004 


2009 


2014 


2018 


2023 


2028 


2032 


2037 





1 


1 


2 


2 


3 


3 


4 


4 


.31 


2042 


2046 


2051 


2056 


2061 


2065 


2070 


2075 


2080 


2084 





1 


1 


2 


2 


3 


3 


4 


4 


.32 


2089 


2094 


2099 


2104 


2109 


2113 


2118 


2123 


2128 


2133 





1 


1 


2 


2 


3 


3 


4 


4 


.33 


2138 


2143 


2148 


2153 


2158 


2163 


2168 


2173 


2178 


2183 





1 


1 


2 


2 


3 


3 


4 


4 


.34 


2188 


2193 


2198 


2203 


2208 


2213 


2218 


2223 


2228 


2234 


1 


1 


2 


2 


3 


3 


4 


4 


5 


.35 


2239 


2244 


2249 


2254 


2259 


2265 


2270 


2275 


2280 


2286 


1 


1 


2 


2 


3 


3 


4 


4 


5 


.36 


2291 


2296 


2301 


2307 


2312 


2317 


2323 


2328 


2333 


2339 


1 


1 


2 


2 


3 


3 


4 


4 


5 


.37 


2344 


2350 


2355 


2360 


2366 


2371 


2377 


2382 


2388 


2393 


1 


1 


2 


2 


3 


3 


4 


4 


5 


.38 


2399 


2404 


2410 


2415 


2421 


2427 


2432 


2438 


2443 


2449 


1 


1 


2 


2 


3 


3 


4 


4 


S 


.39 


2455 


2460 


2466 


2472 


2477 


2483 


2489 


2495 


2500 


2506 


1 


1 


2 


2 


3 


3 


4 


5 


5 


.40 


2512 


2518 


2523 


2529 


2535 


2541 


2547 


2553 


2559 


2564 


1 


1 


2 


2 


3 


4 


4 


5 


5 


.41 


2570 


2576 


2582 


2588 


2594 


2600 


2606 


2612 


2618 


2624 


1 


1 


2 


2 


3 


4 


4 


5 


5 


.42 


2630 


2636 


2642 


2649 


2655 


2661 


2667 


2673 


2679 


2685 


1 


1 


2 


2 


3 


4 


4 


5 


6 


.43 


2692 


2698 


2704 


2710 


2716 


2723 


2729 


2735 


2742 


2748 


1 


1 


2 


3 


3 


4 


4 


5 


6 


.44 


2754 


2761 


2767 


2773 


2780 


2786 


2793 


2799 


2805 


2812 


1 


1 


2 


3 


3 


4 


4 


5 


6 


.45 


2818 


2825 


2831 


2838 


2844 


2851 


2858 


2864 


2871 


2877 


1 


1 


2 


3 


3 


4 


5 


5 


6 


.46 


2884 


2891 


2897 


2904 


2911 


2917 


2924 


2931 


2938 


2944 


1 


1 


2 


3 


3 


4 


5 


5 


6 


.47 


2951 


2958 


2965 


2972 


2979 


2985 


2992 


2999 


3006 


3013 


1 


1 


2 


3 


3 


4 


5 


5 


6 


.48 


3020 


3027 


3034 


3041 


3048 


3055 


3062 


3069 


3076 


3083 


1 


1 


2 


3 


4 


4 


5 


6 


6 


.49 


3090 


3097 


3105 


3112 


3119 


3126 


3133 


3141 


3148 


3155 


1 


1 


2 


3 


4 


4 


5 


6 


6 



APPENDIX 
ANTILOGABITHMS. (Continued?) 



381 



Logarithms 





1 


2 


3 


4 


5 


6 


7 


8 


9 


Proportional parts 


1 


2 


3 


4 


5 


6 


7 


8 


9 


.50 


3162 


3170 


3177 


3184 


3192 


3199 


3206 


3214 


3221 


3228 


1 


1 


2 


3 


4 


4 


5 


6 


7 


.51 


3236 


3243 


3251 


3258 


3266 


3273 


3281 


3289 


3296 


3304 


1 


2 


2 


3 


4 


5 


5 


6 


7 


.52 


3311 


3319 


3327 


3334 


3342 


3350 


3357 


3365 


3373 


3381 


1 


2 


2 


3 


4 


5 


5 


6 


7 


.53 


3388 


3396 


3404 


3412 


3420 


3428 


3436 


3443 


3451 


3459 


1 


2 


2 


3 


4 


5 


6 


6 


7 


.54 


3467 


3475 


3483 


3491 


3499 


3508 


3516 


3524 


3532 


3540 


1 


2 


2 


3 


4 


5 


6 


6 


7 


.55 


3548 


3556 


3565 


3573 


3581 


3589 


3597 


3606 


3614 


3622 


1 


2 


2 


3 


4 


5 


6 


7 


7 


.56 


3631 


3639 


3648 


3656 


3664 


3673 


3681 


3690 


3698 


3707 


1 


2 


3 


3 


4 


5 


6 


7 


8 


.57 


3715 


3724 


3733 


3741 


3750 


3758 


3767 


3776 


3784 


3793 


1 


2 


3 


3 


4 


5 


6 


7 


8 


.58 


3802 


3811 


3819 


3828 


3837 


3846 


3855 


3864 


3873 


3882 


1 


2 


3 


4 


4 


5 


6 


7 


8 


.59 


3890 


3899 


3908 


3917 


3926 


3936 


3945 


3954 


3963 


3972 


1 


2 


3 


4 


5 


5 


6 


7 


8 


.60 


3981 


3990 


3999 


4009 


4018 


4027 


4036 


4046 


4055 


4064 


1 


2 


3 


4 


5 


6 


6 


7 


8 


.61 


4074 


4083 


4093 


4102 


4111 


4121 


4130 


4140 


4150 


4159 


1 


2 


3 


4 


5 


6 


7 


8 


9 


.62 


4169 


4178 


4188 


4198 


4207 


4217 


4227 


4236 


4246 


4256 


1 


2 


3 


4 


5 


6 


7 


8 


9 


.63 


4266 


4276 


4285 


4295 


4305 


4315 


4325 


4335 


4345 


4355 


1 


2 


3 


4 


5 


6 


7 


8 


9 


.64 


4365 


4375 


4385 


4395 


4406 


4416 


4426 


4436 


4446 


4457 


1 


2 


3 


4 


5 


6 


7 


8 


9 


.65 


4467 


4477 


4487 


4498 


4508 


4519 


4529 


4539 


4550 


4560 


1 


2 


3 


4 


5 


6 


7 


8 


9 


.66 


4571 


4581 


4592 


4603 


4613 


4624 


4634 


4645 


4656 


4667 


1 


2 


3 


4 


5 


6 


7 


9 


10 


.67 


4677 


4688 


4699 


4710 


472I| 


4732 


4742 


4753 


4764 


4775 


1 


2 


3 


4 


5 


7 


8 


9 


10 


.68 


4786 


4797 


4808 


4819 


4831 


4842 


4853 


4864 


4875 


4887 


1 


2 


3 


4 


6 


7 


8 


9 


10 


.69, 


4898 


4909 


4920 


4932 


4943 


4955 


4966 


4977 


4989 


5000 


1 


2 


3 


5 


6 


7 


8 


9 


10 


.70 


5012 


5023 


5035 


5047 


5058 


5070 


5082 


5093 


5105 


5117 


1 


2 


4 


5 


6 


7 


8 


9 


11 


.71 


5129 


5140 


5152 


5164 


5176 


5188 


5200 


5212 


5224 


5236 


1 


2 


4 


5 


6 


7 


8 


10 


11 


.72 


5248 


5260 


5272 


5284 


5297 


5309 


5321 


5333 


5346 


5358 


1 


2 


4 


5 


6 


7 


9 


10 


11 


.73 


5370 


5383 


5395 


5408 


5420 


5433 


5445 


5458 


5470 


5483 


1 


3 


4 


5 


6 


8 


9 


10 


11 


.74 


5495 


5508 


5521 


5534 


5546 


5559 


5572 


5585 


5598 


5610 


1 


3 


4 


5 


6 


8 


9 


10 


12 


.75 


5623 


5636 


5649 


5662 


5675 


5689 


5702 


5715 


5728 


5741 




3 


4 


5 


7 


8 


9 


10 


12 


.76 


5754 


5768 


5781 


5794 


5808 


5821 


5834 


5848 


5861 


5876 




3 


4 


5 


7 


8 


9 


11 


12 


.77 


5888 


5902 


5916 


5929 


5943 


5957 


5970 


5984 


5998 


6012 




3 


4 


5 


7 


8 


10 


11 


12 


.78 


6026 


6039 


6053 


6067 


6081 


6095 


6109 


6124 


6138 


6152 




3 


4 


6 


7 


8 


10 


11 


13 


.79 


6166 


6180 


6194 


6209 


6223 


6237 


6252 


6266 


6281 


6295 




3 


4 


6 


7 


9 


10 


11 


13 


.80 


6310 


6324 


6339 


6353 


6368 


6383 


6397 


6412 


6427 


6442 


1 


3 


4 


6 


7 


9 


10 


12 


13 


.81 


6457 


6471 


6486 


6501 


6516 


6531 


6546 


6561 


6577 


6592 


2 


3 


5 


6 


8 


9 


11 


12 


14 


.82 


6607 


6622 


6637 


6653 


6668 


6683 


6699 


6714 


6730 


6745 


2 


3 


5 


6 


8 


9 


11 


12 


14 


.83 


6761 


6776 


6792 


6808 


6823 


6839 


6855 


6871 


6887 


6902 


2 


3 


5 


6 


8 


9 


11 


13 


14 


.84 


6918 


6934 


6950 


6966 


6982 


6998 


7015 


7031 


7047 


7063 


2 


3 


5 


6 


8 


10 


11 


13 


15 


.85 


7079 


7096 


7112 


7129 


7145 


7161 


7178 


7194 


7211 


7228 


2 


3 


5 


7 


8 


10 


12 


13 


15 


.86 


7244 


7261 


7278 


7295 


7311 


7328 


7345 


7362 


7379 


7396 


2 


3 


5 


7 


8 


10 


12 


13 


15 


.87 


7413 


7430 


7447 


7464 


7482 


7499 


7516 


7534 


7551 


7568 


2 


3 


5 


7 


9 


10 


12 


14 


16 


.88 


7586 


7603 


7621 


7638 


7656 


7674 


7691 


7709 


7727 


7745 


2 


4 


5 


7 


9 


11 


12 


14 


16 


.89 


7762 


7780 


7798 


7816 


7834 


7852 


7870 


7889 


7907 


7925 


2 


4 


5 


7 


9 


11 


13 


14 


16 


.90 


7943 


7962 


7980 


7998 


8017 


8035 


8054 


8072 


8091 


8110 


2 


4 


6 


7 


9 


11 


13 


15 


17 


.91 


8128 


8147 


8166 


8185 


8204 


8222 


8241 


8260 


8279 


8299 


2 


4 


6 


8 


9 


11 


13 


15 


17 


.92 


8318 


8337 


8356 


8375 


8395 


8414 


8433 


8453 


8472 


8492 


2 


4 


6 


8 


10 


12 


14 


15 


17 


.93 


8511 


8531 


8551 


8570 


8590 


8610 


8630 


8650 


8670 


8690 


2 


4 


6 


8 


10 


12 


14 


16 


18 


.94 


8710 


8730 


8750 


8770 


8790 


8810 


8831 


8851 


8872 


8892 


2 


4 


6 


8 


10 


12 


14 


16 


18 


.95 


8913 


8933 


8954 


8974 


8995 


9016 


9036 


9057 


9078 


9099 


2 


4 


6 


8 


10 


12 


15 


17 


19 


.96 


9120 


9141 


9162 


9183 


9204 


9226 


9247 


9268 


9290 


9311 


2 


4 


6 


8 


11 


13 


15 


17 


19 


.97 


9333 


9354 


9376 


9397 


9419 


9441 


9462 


9484 


9506 


9528 


2 


4 


7 


9 


11 


13 


15 


17 


20 


.98 


9550 


9572 


9594 


9616 


9638 


9661 


9683 


9705 


9727 


9750 


2 


4 


7 


9 


11 


13 


16 


18 


20 


.99 


9772 


9795 


9817 


9840 


9863 


9886 


9908 


9931 


9954 


9977 


2 


5 


7 


9 


11 


14 


16 


18 


20 



382 



CALCULATIONS OF ANALYTICAL CHEMISTRY 



INTERNATIONAL ATOMIC WEIGHTS 
(1944) 





Sym- 
bol 


At, 
num- 
ber 


At. 
wt. 




Sym- 
bol 


At. 
num- 
ber 


At. 
wt. 


.A. In min u m 


Al 


13 


26.97 


Molybdenum 


Mo 


42 


95.95 


Antimony 


Sb 


51 


121.76 


Neodyrnium 


Nd 


60 


144.27 


Arcron 


A 


18 


39.944 




Ne 


10 


20.183 




As 


33 


74.91 


Nickel 


Ni 


28 


58.69 


Barium 


Ba 


56 


137.36 




N 


7 


14.008 


Beryllium 


Be 


4 


9.02 


Osmium 


Os 


76 


190.2 


Bismuth 


Bi 


83 


209 00 


Oxygen 


o 


8 


16.0000 


Boron 


B 


5 


10.82 


Palladium 


Pd 


46 


106.7 


Bromine 


Br 


35 


79.916 


Phosphorus 


P 


15 


30.98 


Cadmium 


Cd 


48 


112.41 


Platinum 


Pt 


78 


195.23 


Calcium 


Ca 


20 


40.08 


Potassium 


K 


19 


39.096 


Carbon 


c 


6 


12.010 


Praseodymium 


Pr 


59 


140.92 


Cerium . 


Ce 


58 


140.13 


Protactinium 


Pa 


91 


231 




Cs 


55 


132.91 


Radium 


Ra 


88 


226.05 


Chlorine? 


Cl 


17 


35.457 


Radon 


Rn 


86 


222 


Chromium 


Cr 


24 


52.01 


Rhenium 


Re 


75 


186.31 


f*nVmlt 


Co 


27 


58 94 


Rhodium 


Rh 


45 


102.91 


Columbium 


Cb 


41 


92.91 


Rubidium 


Rb 


37 


85.48 


CoDDor 


Cu 


29 


63.57 


Ruthenium 


Ru 


44 


101.7 




Dv 


66 


162 46 


Samarium 


Sm 


62 


150.43 


T^rhin m 


Er 


68 


167.2 




Sc 


21 


45.10 




Eu 


63 


152 


Selenium 


Se 


34 


78. 9f, 


Fluorine* 


F 


9 


19.00 




Si 


14 


28.06 


Gadolinium 


Gd 


64 


156 9 


Silver 


Ag 


47 


107.880 


Gallium 


Ga 


31 


69.72 


Sodium 


Na 


11 


22.997 


Germanium 


Ge 


32 


72.60 


Strontium 


Sr 


38 


87.63 


Gold 


Au 


79 


197.2 


Sulfur 


S 


16 


32.06 


Hafnium 


Hf 


72 


178.6 


Tantalum 


Ta 


73 


180.88 


Helium 


He 


2 


4.003 


Tellurium 


Te 


52 


127.61 




Ho 


67 


164.94 


Terbium 


Tb 


65 


159.2 


TT A 


H 


1 


1.0080 


Thallium 


Tl 


81 


204.39 


TtlHlllTn 


In 


49 


114.76 


Thorium 


Th 


90 


232.12 


Iodine 


I 


53 


126.92 


Thulium 


Tm 


69 


169.4 




Ir 


77 


193.1 


Tin 


Sn 


50 


118.70 


T 


Fe 


26 


55.85 


Titanium 


Ti 


22 


47.90 


ICrvrjton 


Kr 


36 


83.7 


Tungsten 


W 


74 


183.92 




T a 


57 


138 92 


Uranium . 


u 


92 


238 14 


Lead 


Pb 


82 


207.21 


Vanadium 


V 


23 


50.95 


T ittiiiilYl 


Li 


3 


6.940 


Xenon 


Xe 


54 


131.3 


Lutecium 


Lu 


71 


174.99 


Ytterbium 


Yb 


70 


173.04 




Me 


12 


24.32 


Yttrium 


Y 


39 


88.92 




Mn 


25 


54.93 


Zinc 


Zn 


30 


65.38 




He 


80 


200.61 


Zirconium 


Zr 


40 


91.22 



















INDEX 



Absolute error, 2 
Absorbing agents, 287 
Absorption methods, 286 
Accuracy of a result, 2 
Acid mixtures, titration of, 184 
Acidimetry, calculations of, 158 
Acidity, control of, in sulfide precipi- 
tations, 68 
Activity, 55 
Activity coefficients, 55 
Adsorption indicators, 243 
Alkali group, problems on, 312 
Alkalimetry, calculations of, 158 
Alkaline earth group, problems on, 

312 

Aluminum, problems on, 317 
Ammonium, problems on, 314 
Ammonium sulfide group, problems 

on, 311 

Ampere, definition of, 126 
Amperometric titrations, 275 
Anion group, problems on, 313 
Antilogarithm, definition of, 12 
Antimony, problems on, 323 
Arsenic, problems on, 323 
Atomic weights, calculation of, 109 
Average deviation, 3 
Avogadro's law, 281 

B 

Balance, sensitiveness of, 93 
Barium, problems on, 316 
Beryllium, problems on, 320 
Bismuth, problems on, 320 
Bismuthate method, for manganese, 

222 
Boiling point, raising of, 142 



Boron, problems on, 320 
Boyle's law, 279 
Brass, problems on, 321 
Bromine, problems on, 315 
Bronze, problems on, 323 
Buffered solution, 52 
Buffered solutions, applications of, in 
separations, 67 



Cadmium, problems on, 321 
Calcium, problems on, 316 
Calibration of measuring instru- 
ments, 153 

Calibration corrections, table of, 98 
Calomel cell, 255 
Carbon, problems on, 324 
Carbon dioxide, problems on, 324 
Carbonate mixtures, titration of, 201 
Cement, problems on, 317 
Ceric sulfate process, 226 
Cerium, problems on, 320 
Characteristic, definition of, 11 
Charles's law, 280 

Chemical balance, sensitiveness of, 93 
Chemical equations, ionic, rules for 
writing, 18 

mathematical significance of, 29 

oxidation-reduction, rules for writ- 
ing, 21 

purpose of, 16 

types of, 16 
Chemical factor, 104 
Chemical formula, mathematical sig- 
nificance of, 28 

Chlorate method, for manganese, 223 
Chlorine, problems on, 315 
Chromium, problems on, 325 
Cobalt, problems on, 329 



383 



384 CALCULATIONS OF ANALYTICAL CHEMISTRY 



Cologarithm, definition of, 11 
Combustion methods, 287 
Common ion effect, 52 
Complex-ion formation, separations 

based on, 69 

Complex-ion-formation methods, 249 
Complex ions, dissociation constants 

of, 56 

Concentration, methods of express- 
ing, 36 

of solutions, 36 
Concentration cell, 85 
Conductance, definition of, 266 
Conductometric titrations, in acidim- 
etry, 267 

apparatus for, 271 

general discussion of, 266 

in precipitimetry, 270 
Copper, problems on, 321 
Coulomb, definition of, 126 
Cubic centimeter, definition of, 153 
Cumulative corrections, 99 
Current efficiency, 129 
Cyanide, problems on, 315 

volumetric determination of, 249 

D 

Dalton's law, 280 

Data, conversion to milliequivalents, 

169 

Decomposition potential, 125 
Definite proportions, law of, 102 
Densities, table of, 96 
Deviation measure, 2 
Dichromate process, 225 
Diffusion current, 275 
Digit, definition of, 5 
Distribution law, 73 
Distribution ratio, 73 
Double-indicator titrations, 199 

E 

Electrode potential, relation to con- 
centration, 83 
Electrode potentials, 78 



Electrolysis, analysis by, 126 
Electrolytic methods, 125 
Electrometric methods, 255 
Elimination of a constituent, 135 
Empirical formulas, calculation of, 

144 

End point, definition of, 191 
Equations, for half-cell reactions, 

rules for writing, 79 
ionic, rules for writing, 18 
mathematical significance of, 29 
oxidation-reduction, rules for writ- 
ing, 21 

purpose of, 16 
types of, 16 
Equilibrium constant, calculation of, 

from electrode potentials, 87 
Equilibrium constants, 46 
Equivalence point, definition of, 190 

determination of pH at, 191 
Equivalent conductance, definition 

of, 266 

Equivalent weight, definition of, 38 
Equivalent weights, in complex ion 

methods, 249 

in neutralization methods, 158 
in oxidation-reduction methods, 

211 

in precipitation methods, 243 
Erg, definition of, 130 
Evolution method, for sulfur, 236 



Factor weight sample, 110 
Faraday, definition of, 127 
Faraday's laws, 126 
Ferrous sulfate, equivalent weight of, 

212 

Fluorine, problems on, 315 
Formal solution, definition of, 37 
Formula weight, definition of, 28 
Formulas, calculations based on, 28 
Fractional precipitation, 62 
Freezing point, lowering, 142 
Fuming sulfuric acid, titration of, 

184 



INDEX 



385 



G 

Gas absorption methods, 286 

Gas analysis, absorbing agents in, 

287 

calculations of, 279 
divisions of, 281 
laws of, 279 

Gas combustion methods, 287 
Gas-volumetric analysis, calculations 

of, 283 

Gas-volumetric methods, 281 
Gay-Lussac's law, 281 
General analysis, problems on, 335 
Glass electrode, 258 
Gold, problems on, 315 
Gram-atom, definition of, 28 
Gram-equivalent weight, definition 

of, 38 

Gram-ion, definition of, 28 
Gram-molecular weight, definition 

of, 28 



H 

Half cell, 77 

Half-cell reactions, rules for writing, 
79 

Halogens, problems on, 315 

Hydrogen electrode, 78 

Hydrogen peroxide, equivalent 
weight of, 214, 216 

Hydrogen sulfide, equivalent weight 
of, 214 

Hydrogen sulfide group, problems on, 
310 

Hydrogen sulfide precipitation, con- 
trol of acidity in, 68 

Hydrolysis, calculation of degree of, 
196 

Hydronium ion, 47 



Indicator constant, 189 
Indicators, adsorption, 243 
analyses involving two, 199 



Indicators, properties of, 188 

table of, 190 
Indirect methods, gravimetric, 118 

volumetric, 176 

Introduction of a constituent, 135 
lodimetric process, 234 
Iodine, equivalent weight of, 216 

problems on, 315 
Ionic equations, method of writing, 18 

oxidation-reduction, method of 

writing, 21 
lonization of acids, bases, and salts, 

17 

lonization constant, 50 
Iron, problems on, 317 
Iron value, 225 
Isomorphic replacement, 145 



Joule, definition of, 130 



K 
Kjeldahl method, for nitrogen, 176 



Lead, problems on, 321 

Liebig method, for cyanide, 249 

Lime, problems on, 317 

Limestone, problems on, 317 

Liter, definition of, 153 

Logarithm tables, method of using, 13 

Logarithms, rules governing use of, 9 

M 

Magnesium, problems on, 316 
Manganese, problems on, 327 
Mantissa, definition of, 11 
Mass action, law of, 46 
Mathematical operations, 1 
Mean deviation, 3 
Mean value, 2 

Measuring instruments, calibration 
of, 153 



386 



CALCULATIONS OF ANALYTICAL CHEMISTRY 



Mercury, problems on, 315 

Milliequivalent weight, definition of, 
38 

Milliliter, definition of, 153 

Millimole, definition of, 29 

Mixed alkali titration, with two in- 
dicators, 199 

Mixtures, determination of com- 
ponents in, 183 

Mobility of ions, 266 

Molar solution, definition of, 37 

Mole, definition of, 28 

Molecular formulas, calculation of, 
141 

Molybdenum, problems on, 324 

N 

Neutral point, definition of, 191 
Neutralization methods, 158 
Nickel, problems on, 329 

volumetric determination of, 251 
Nitrogen, problems on, 314 
Normal solution, definition of, 38 
Normal temperature, definition of, 

153 
Normality, adjustment of, 163 

determination of, 168 

of mixed solutions, 161 
Number, definition of, 3 
Numerical problems, conventions re- 
garding, 6 



O 

Ohm, definition of, 130 

Ohm's law, 130 

Oleum, titration of, 184 

Oxalic acid, equivalent weight of, 213 

Oxidation, definition of, 211 

Oxidation number, 20 

Oxidation potentials, 76 

Oxidation-reduction, relation of cur- 
rent to, 76 

Oxidation-reduction equations, in 

terms of half-cell reactions, 80 
method of writing, 21 



Oxidation-reduction methods, 211 

Oxidation-reduction processes, cal- 
culation of, 216 

Oxidation-reduction reactions, cal- 
culation of extent of, 86 

Oxidimetry, 211 

Oxidizing agents, equivalent weights 
of, 215 



Percentage purity, calculation of, 

from titration values, 173 
Percentages, calculation of, 105 

calculations from, 135 
Permanganate process, 219 
Permanganate titrations, in neutral 

solution, 223 
Persulfate method, for manganese, 

223 

Phosphate mixtures, titration of, 207 
Phosphorus, problems on, 329 
pll value, definition of, 48 

determination of, at equivalence 

point, 191 

Platinum, problems on, 315 
pOH value, definition of, 48 
Polarograph, 277 

Polybasic acids, ionization of, 18, 53 
Potassium, problems on, 313 
Potassium binoxalate, equivalent 

weight of, 213 
Potassium bromate, equivalent weight 

of, 216 
Potassium dichromatc, equivalent 

weight of, 215 
Potassium ferricyanide, equivalent 

weight of, 215 
Potassium permanganate, equivalent 

weight of, 215 
Potassium tetroxalate, equivalent 

weight of, 214 

Potentiometric methods, 255 
Potentiometric titrations, in acidim- 

etry, 255 

apparatus for, 256 
in oxidimetry, 259 
in precipitimetry, 261 



INDEX 



387 



Precipitation methods, 243 

Precision measure, 2 

Problems, conventions regarding, 6 

Q 

Quinhydrone electrode, 257 

R 

Range of doubt, 189 

Reacting solutions, ratios of, 166 

Redox methods, 211 

Reducing agents, equivalent weights 

of, 212 

Reduction, definition of, 211 
Relative error, 2 
Reliability, factors influencing, 1 
Reported percentages, calculations 

from, 135 

S 

Selenium, problems on, 331 

Sensitiveness of a balance, 93 

Significant figure, definition of, 3 

Significant figures, rules governing 
use of, 3 

Silicon, problems on, 324 

Silver, problems on, 315 

Silver group, problems on, 309 

Slide rule, use of, 14 

Smith, J. L. method, for alkalies, 120 

Sodium, problems on, 313 

Sodium carbonate, titration of, 199 

Sodium thiosulf ate, equivalent weight 
of, 214 

Solubility product, 60 

Solutions, adjusting to desired nor- 
mality, 163 

methods of standardizing, 168 
normality of mixed, 161 

Specific conductance, definition of, 
266 

Specific electrode potentials, 76 

Specific elements, methods for, 296 
problems on, 309 

Stannous chloride, equivalent weight 
of, 213 

Strontium, problems on, 316 



Substitution, method of, 97 
Sulfur, problems on, 331 
Swings, method of, 94 



Thiocyanate, problems on, 315 
Thorium, problems on, 320 
Tin, problems on, 323 
Titanium, problems on, 317 
True volume, calculation of, 153 
Tungsten, problems on, 324 
Two indicators, titrations involving 
use of, 199 

U 
Uranium, problems on, 320 



Vacuo, conversion to weight in, 95 
Vanadium, problems on, 325 
Vapor pressure, correction for, 282 
Volhard method, for halides, 244 

for manganese, 223 

for Silver, 244 
Volt, definition of, 130 
Volume, of reagent, calculation of, 
113 

relationship to normality, 166 

relationship to percentage, 180 
Volume ratios, 37 
Volumetric analysis, divisions of, 158 

W 

Water, ion product constant of, 47 

problems on, 313 
Water vapor, correction for, 282 
Watt, definition of, 130 
Weight, conversion to vacuo, 95 
Weights, calibration of, 97 
Williams method, for manganese, 223 

Z 

Zinc, problems on, 321 
Zirconium, problems on, 320